Last Minute Revision Series - CA Foundation M'26 - Permutations & Combinations | Prof. Aman Khedia

Added: 2026-05-09

[00:00:00]Okay. Hi everyone guys. Now let's start with the practice of the chapter permutations and combination. I hope you are done with the revision. Now this is the right time to start the practice session of the chapter PNC. Let's start with the first variety problems based on word. We are going to cover all the variety all the kind of questions miscellaneous question everything will remain untouched. Let's start guys. In how many ways the letters of the word accountant be arranged so that vowels are always together. Let's put the vowels in one box. How many vowels are there? A o u a o u a. Let's put them in one box. Now what is left guys? Double c o done. U done N T A is done N T guys let's do the counting everybody 1 2 3 4 5 6 7 guys 7 factorial but when you done the seven factorial you made one mistake you violated the conditions of the logic number one you have arranged 2 C divided by 2 factorial you have arranged 2 N again divide by 2 factorial you have arranged 2 T again divided by two factorial. But Aman if we apply arrangement inside the box still vowels are together. How many wles are there? How many things are there inside the box? You need to arrange four factorial. But when you done four factorial again you violated the condition you arrange 2 a again divided by 2 factorial. Guys, hold your calculator. 5 4 7 factorial 7 into 6 into 5 into 4 into 3 into 2 into 1 5 0 4 0 * 4 factorial 24 divided by 2 divided by 2 divided by 2 divide by 2 again. So the correct answer is 7560 everybody guys with me 7 560 option B is the right answer. Sorry 7560 option A is the correct answer guys everybody you have to solve with me. Try to have more speed than me or match your speed with me but speed should not be less than with my discussion. Now in how many ways the letters of the word arrange be arranged. So guys 1 2 3 4 5 6 1 2 3 4 5 6 7 There are seven things 7 p 7 factorial but guys don't you think you have reached 2 a divide by 2 factorial everybody? Now Aman sir we have arranged 2 R again divide by 2 factorial. Are there any more identicals? No. Again 5 0 4 0 divided by 2 divided by 2 1 2 6 0.

[00:02:49]Everybody guys 1 2 6 0 option C is the right answer. In how many ways letters of the word strange be arranged so that vowels never come together. How to solve never come together? Ammons are total ways minus ways when they come together. Guys, in how many ways the letters of the word strange be arranged? 1 2 3 4 5 6 7 factorial. Is there any identical objects in the seven factorial? No. All are distinct objects. Now divided by guys divided by everybody with me minus when vowels are together guys minus how many vowels are there a and e let's put a and e in one box what is left s t r n g let's do the counting 1 2 3 4 5 6 minus when they come together 6 factorial even if they are arranged inside the box still they are together so the answer is 7 factorial 5040 - 720 * 2 I'm answer 1 1440 the answer will be guys uh 5040 minus this answer will be guys 3,600 everybody option A is the right answer everybody guys option A is the right Answer guys. Yes, everybody. Very good.

[00:04:34]Question number four.

[00:04:36]The number of combinations of the word college taken four at a time. Guys, can you tell me which logic will apply? Think guys, which logic will apply. Mark this question as an LDR.

[00:04:50]Just mark this question as an LDR.

[00:04:54]Why? Find the number of combinations of the letters of the word college. Ammon, how many alphabets are? 1 2 3 4 5 6 7.

[00:05:03]There are seven items. You have to select only four. Condition one.

[00:05:08]Condition two. And yes, there are identicals. This is a based on the logic number five of the problems based on word. Why? See guys, objects are many. You have to select only four and yes there are identicals.

[00:05:23]So there will be cases will be there.

[00:05:25]Ammon sir, how many identical objects are there guys? You have 2 L then you have two E then guys you have C O G C O G am sir we need to make a word of four only. Now how many things can we formed guys? Case number one am two alike of one kind two alike of another guys. Case number two. Very good guys.

[00:05:54]Who has said this answer? Arjun. Very good. Arjun. Arjun. Very good.

[00:05:58]Impressive answer guys. I'm answer two alike.

[00:06:03]Two different.

[00:06:04]Case number three guys. All four different. All four different. Ammon sir. Guys they have simply said the word combination. It means there is no need of arrangement. Only I have to do the selection. I need two alike. I need two alike and two alike. Out of this two pair, if both the pairs come, can I say I will have two alike and two alike. Out of two pairs, I need entire two pairs.

[00:06:30]Out of two, I need to select to 2 C2.

[00:06:33]Now two alike, out of these two pair, if any one pair will come, I will have two alike. 2 C1 and multiply guys. Let us assume one pair has taken place. Now two different how many we have? 1 2 3 2 E cannot come 1 E can come four out of four you need two multiply by 4 C2 all four different guys all four different 1 2 3 4 5 out of five I need any four five C 4 let's do it guys 2 C2 is 1 this 2 C2 is 1 4 C2 4 into 3 / * 2 guys you will get this is out outcome is 12 5 C4 5 into sorry 4 into 3 divid by 2 yes sorry this outcome will be guys not uh uh yes 6 into 2 yes 5 C4 5 into 4 into 3 into 2 divided by 24 everybody you are having this one is five so guys how much is the total number of ways sir 12 + 5 12 + 5 + 1 answer 18 ways 18 ways guys we have solved one question in the class in how many ways the fourletter word can be formed from the letters of the word mathematics these are same question guys here word mathematics has been replaced by the word college everybody is this point clear to all of you guys can I move forward everybody is this question clear to all of you yes balam 18 is the right answer now guys this is a question of the number Mistakely wise this question is here but this is a very beautiful question. So let's discuss find the number of even number first condition I need to form a even numbers greater than 100 using this digits am when we say the number is an even when the last place of the term is either divisible by two or it ends on zero. So guys here if the zero comes in the end it will be an even number we have two also two is uh the number which can make the number even if it ends on two again we have a number even number they are saying greater than 100 greater than 100 it means three digit but they are not restricting us whatever number of numbers you can make greater than 300 we are allowed ammon you are having four digits it means even we can make four-digit number also three-digit Greater than 100 means five digits, six digit. But why we will not make five digit and six-digit number? Because we are not having that much number of digits. We are having maximum four digits. It means we can form maximum fourdigit number. Amster now let's split the cases. Let's split the cases. Let's solve three digits.

[00:09:32]Let's solve four digits here guys in three digits. 1 2 3 1 2 3 Why we are splitting the case? Because if it ends on zero, then also it is even and if it ends on two, then also it is even. Guys, yes, zero is having the possibility that it will never come in the first place. So guys, there is a two place restriction. On a safer side, I will split the case. Same I will do it for the four digit 1 2 3 4 1 2 3 4. What if it ends on zero? What if it ends on two? Amen. What if it ends on zero? Here only one number can come that is zero.

[00:10:12]Now if the zero has taken place. Now what whatever it starts it will be always greater than 100. How many numbers are there to make it greater than 100? All the three numbers. If any number one has taken place now how many numbers can can come two numbers outcome is six. Amen sir. If the it ends on two again one number has taken place and the number which has taken place is two. Now here you here now zero cannot come. Now here the zero cannot come because if the zero will come the number will be two digit. Now zero cannot come. Now how many numbers can come? I'm answer any two number can come. Let's say any one number has taken place. Now it even the two can come. So how many numbers can come? 2 into 2 into four guys how much?

[00:10:57]Four. So guys how many three-digit number can be formed even greater than 100 6 + 4 10 but they have said greater than 100 even if you are make if you are capable of making fourdigit five digit you can make but why we are only considering till fourdigit because we are not having numbers we are having only four digits so I will make only up to four digit numbers now what if it ends on two zero one number can come and the number which can come is zero now guys how many numbers can come here.

[00:11:28]Amen. Sir, whatever comes it will be always a greater than 100. Sir, three numbers can come. Then here guys, two numbers can come again. Guys, here one number can come. Multiply guys, you will get six. What if it ends on two? A man s if it ends on two. Now here zero cannot come. Why zero cannot come? Because of the zero will come here. The number will be three-digit. And I have already considered three-digit number. Zero cannot come. Now again two numbers can come. If any number has taken place now zero is capable of coming again two numbers can come and here guys one number can come if you multiply them again you will get four so the total number will be 10 + 10 20 ways guys 10 + 10 20 ways everybody guys is this question clear guys a very good question everybody Arjun Balam uh Mon everybody Can I move forward everybody? Guys, can I move forward with this?

[00:12:34]Yes. Now guys, make a correction over here. If this is not correct in your book, guys, algebra spelling must check.

[00:12:41]The spelling must be A L G E B R A, guys. A L G E B R A guys. A L G E B R A guys. A L G E B R A guys. Everybody with me now. How many ways the letters of the word algebra be arranged without changing the relative orders of the vovel? without changing the relative orders of the vowel. Guys, let's draw the places. How many? 1 2 3 4 5 6 7.

[00:13:26]Guys, 1 2 3 4 5 6 7. They are saying algebra must occupy the same place which they were occupying originally. Amir, how many algebra? A, E and A. The first place must be occupied by the algebra.

[00:13:44]Sorry. Then 1 2 3 4. Fourth place 1 2 3 4. Fourth place must be occupied by algebra and the last place must be occupied by the algebra. This says guys that these three places must be occupied by algebra. Now guys, I am not saying that in the first place only a will come. But I am saying in first place, fourth place and last place only algebra can come. In these three places any three algebra can come. In how many ways? Three algebra can be arranged in three places. Ammons are 3 P3.

[00:14:17]But when you said 3 P3, you have made one mistake. You have arranged 2 A divided by 2 factorial multiply by now how many? 1 2 3 4. Now guys, how many ways in which four places can be arranged by the four numbers? 1 2 3 4. Again guys, 4 P4.

[00:14:39]So guys 3 P3 is 3 factorial into 4 factorial divided by 2 guys 3 factorial is 6 * 1/4 sorry 6 * 24 divided by 2 everybody guys the answer will be 72 ways guys the answer will be 72 ways.

[00:14:57]Next question guys in how many ways the letters of the word director be arranged so that three vowels are never together.

[00:15:05]Same thing as wobbles are never together. So guys first we will say among the total ways minus ways when they are together total ways minus the ways when they are together guys in how many ways you can arrange the word director 1 2 3 4 5 6 7 8 guys 8 factorial 8 P8 8 factorial but don't you think you have here guys 2 T I R E C T O R here you have two R among intent by mistake I have arranged two identical objects remove so this is how many ways you can arrange the word character minus when the vowels are together let's put the vowels in one box how many E I put E and I E I and O E I and O in one box. Guys, what are left?

[00:16:07]D I is in box. D R E is in box. D R C T R D R D R C T R. Guys, let's count. 1 2 3 4 5 6 guys. 6 factorial. But when you return six factor again, you made one mistake. You have arranged two rs divided by two factorial. But ammon even inside this box if you are in these three things still wobbles are together multiply by 3 factorial. So guys the answer will be 8 factorial 8 into 7 into 6 into 5 into 4 into 3 into 2 into 1 divided by 2 guys 2060 6 factorial 720 * 6 divided by 2 m - mrc that the answer will be 18,000 the answer will be guys 18,000 ways everybody guys the answer will be 18,000 ways everybody guys let's move ahead Okay guys, how many ways how many words can be formed from the letters of the word oriental? Oriental. So at A and E always occupy odd places among the let's draw the places 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 How many odd place are there? 1 3 5 and 7. In this seven places always a and e should take the odd place. So in how many ways I can select out of seven I can select two place and I can arrange two place.

[00:17:55]Seven B2 out of this any place any two place must have gone. Now A and E is also gone. A and E is also gone. Now how many things are left? 1 2 3 4 5 6 In how many ways I can select and arrange six persons in the sixth places? 1 2 3 4 5 6 multiply by amir 6 P 6. Let's do it guys. 6 P 6 is 6 factorial 7 P to 7 into 6 42 * 6 factorial that is multiply by 720. The answer is guys uh sorry sorry my mistake my mistake my mistake guys how many odd place 1 2 3 4 4 P2 sorry guys my my mistake 4 P2 in how many ways I can select any two places out of the four odd places 4 P2 yes yes guys 4 into 3 12 multiply by 720 guys are you all getting 8 6 4 0 8 6 4 0 guys 840 is the right answer guys can Can I move forward everybody? Yes. Can I move forward? Yes.

[00:19:04]In how many ways the letters of the word bharat be arranged so that B and H never come together? Ammon sir again we will do the same thing. Total ways minus when they come together. Total ways minus when they come together. In how many ways total word bharat word can be arranged?

[00:19:28]Let's count guys. 1 2 3 4 5 6 I'm answer six factorial. But don't you think when you are saying six factorial you made one mistake there are two times there is a so you have to divide it by two factorial. Now B and H always come together. So you have to put B and H in one box. How many things are left guys? If you put B and H A R A T A R A T guys in how many ways they come together? Let's count this box as an 1 1 2 3 4 5 factorial. But when you written five factorial again you made the mistake you have arranged this 2 a divided by 2 factorial in how many answer? Even if they arrange inside the box still they are together guys. This will cancel and how much you will get 6 factorial 720 divided by 2 m + 5 factorial 120 m minus I'm answer 240 is the right answer 240 base this is the right answer everybody guys is this point clear to all of you guys is this question clear to all of you can I move forward yes I'm answer now in how many ways if fourlet word are taken with or without meaning from the word logarithm without repetition. How many words can be formed? Am I answer alphabets are many?

[00:21:02]We have to take only four. Should I apply the logic number five? No. This is one condition. Second condition. Are there identicals? Let's check guys. This don't have any uh this don't have any identical. All are different. 1 2 3 4 5 6 7 8 9. They all nine are different. So whenever you take any four it will be always different. So the answer is very simple. In how many ways I can take form four alphabet word out of 9 alphabet?

[00:21:33]9P4 guys. What is 9 P4? 1 2 3 4 sir.

[00:21:38]3024 everybody. 3024. This is the right answer guys. Here this was the right answer guys. Everybody is this point clear to all of you guys? Can I move forward? Is this point clear to all of you?

[00:21:54]Let's move forward. Find the sum of fourdigit number. Okay. How many four is the sum of all the fourdigit number that can be formed from this digits? Guys, I have given you one shortcut. Sum of all the digits. Sum of digits multiply by n minus one factorial multiply by guys multiply by 1 1 1. But if you get identical then divide by such number of factorial. Let's use the calculator. Everybody guys let's use the calculator. Answer. Let's use it. Just a second guys. Just a second guys. Let's do it. Sum of all the digits. Everybody with me? 3 + 4 + 5 + 5 * n minus 1.

[00:22:44]Fourdigit number. So 4 - 1 3 factorial.

[00:22:47]3 factorial is 6 mult* 1 1 1 guys this divided by guys how many identicals are there? Two. Again guys divided by 2 factorial that is 2 you will get 5 61 everybody guys. Tarkesh right answer. So the answer will be option D 5 6 1 5 6 1 guys moving forward how many numbers can be formed with the help of this numbers which is not divisible by five guys a number is divisible by five if it ends on 0 and five zero or five if the five-digit numbers are not repeating there is no repetition is allowed five-digit number 1 2 3 4 5 guys I have to form five-digit number guys there is a restriction in the last place what is the restriction that it should never be divisible by five among sir it means there is a restriction in the last place the last place should not have zero and five rest all the number it can have so I will resolve the place having restriction sir five cannot come rest all the numbers can come how many number can come 1 2 3 4 Five.

[00:24:02]So five numbers can come. Let us say any one number has taken place. Now here there is no restriction. Now the number which has taken place it cannot come.

[00:24:10]But now five is available to come again.

[00:24:12]Five numbers can come. Five 4 3 and two guys. How many numbers will what will the answer everybody? Sir 5 into 4 into 3 into 2 into 5. Guys are you all getting the answer 600?

[00:24:28]Everybody are you all getting the answer 600 ways guys are you sir since we have four digit that's why we are taking 411 yes sir if you are having five digits you have to take 1 one one five times okay now the number of threedigit odd numbers guys number is an odd number if it is not divisible by two number is not divisible by two if it does not if it ends on zero if it ends on two then only it is divisible by two then only it is an even Odd number the last it should not end on Z number which is multiple of two and it should not end on zero. Okay.

[00:25:05]Three digit 1 2 3 guys. It should be a odd number having restriction in the last place. Okay. If it ends on five, it will be an odd. If it ends on seven, it will be an odd. It if it ends on 9, it will be a not by any chance. If it ends on six or eight, it will be an even number. So guys how many numbers can come at the last amas only any three numbers can come let's say nine has taken place now there is no restriction now how many numbers can come 1 2 3 4 here how many numbers can come three guys 4 into 3 into 36 ways 36 ways guys is this point clear to all of Huh?

[00:25:55]So sorry guys. Four guys guys guys. One mistake. Wait wait wait wait wait wait guys wait wait wait wait wait guys guys. This time they have said repetition is allowed. Superb question guys. Mark this question as an LDR guys. Mark this question as an LDR.

[00:26:16]Repetition is allowed. If any number has taken place this time I will not cancel.

[00:26:20]Guys at the end only three numbers can come because the number has to be odd.

[00:26:24]Now the number which has taken place it can repeat. If repetition is allowed guys all the numbers 1 2 3 4 5 all the five numbers are available. Even if any number has taken place it can come again. So again five numbers can come.

[00:26:38]How much is the answer guys? 5 into 5 into 3 everybody guys 75 ways 75 ways is the right answer. Next question guys, how many numbers of three digit can be formed using the digits guys? Three digit number 1 2 3. There is no repetition guys. A single brave question. Now 1 2 3 4 5 here five numbers can come. No restriction. Then there is four numbers can come and here guys three numbers can come. What is guys? Five into 4 into 3 guys 60 ways.

[00:27:15]Everybody 60 ways is the right uh answer. Moving forward now moving forward. Yes. Yes. Yes. Yes. Yes. We moving forward guys. Huh?

[00:27:26]Yes. 60 ways. Why you are saying 120?

[00:27:28]No. No guys. 60 ways. 60 will be the right answer. No digits being repeat.

[00:27:32]Repetition is not allowed. Okay. 1 2 3 4 5. Yes. Now guys, let's solve how the number of numbers between 1,000 and 10,000. Between 1,000 and 10,000 guys after 1,000 what is the next number?

[00:27:48]1,000 and 1,000 and 2,000 and three fourdigit number. Now before 10,000 between before 10,000 number is 99 999.

[00:27:59]This is also four-digit number. In short they are asking us to make a fourdigit number. 1 2 3 4 guys without repetition.

[00:28:08]1 2 3 4 5 6. There is no other restriction. Amas here six numbers can come here five numbers can come here four numbers can come and guys here three numbers can come so guys what will be the answer everybody 6 into 5 into 4 into 3 ways 360 ways everybody option C will be the right answer yes guys option C will be the right answer option C will be the right answer guys now how many seven-digit Number can be formed. No digits being repeated not allowed. Not divisible five. Number should not be divisible by five. Number should not be divisible by five. And repetition is not allowed. Guys this time same question but you have to form a seven-digit number. 1 2 3 4 5 6 7 the number should not be divisible by five. It means here in the end zero cannot come and five cannot come. What if they would have given you zero also here? I would further split the case. I would have What if it ends on zero? What if it ends on five? But since they have not given zero, only one number can make it divisible by five. That is a five. So here five cannot come. Rest all the numbers can come. Let's count how many numbers are there. 1 2 3 4 5 6 am six numbers. If any one number has taken place, now that can number cannot come.

[00:29:39]But now the five is available. Now the five is eligible to come. Now how many numbers can come again? Six. Then five 4 3 2 1. Let's multiply them.

[00:29:55]Let's multiply them. 6 into 5 into 4 into 3 into 2 into 1. Multiply by 6. Are you all getting 4320 is yes. So guys 4320 everybody option B is the right answer guys. And here we are done with the practice of the problems based on word problems based on number. Moving forward guys.

[00:30:20]Now problems based on geometry guys everybody remembers that problem based on geometry guys here guys how many guys they have given you number of diagonals.

[00:30:31]First of all make a correction guys.

[00:30:33]Here they should have given the option 11 not 9. Guys this question has a correction 11. Now am sir there are 44 diagonals then the number of sides. See guys let me make you revise once. If you have the rectangle how many diagonals can be made? Amir two diagonals can be made.

[00:30:54]But to draw the diagonal what is your requirement? Amir our requirement is four corner point out of four you need any two corner point I need two corner point out of four when you solve 4 C2 you will get six but practically we can get only two diagonals why we are getting six guys you will get two diagonals and you will get four sides 1 2 3 4 So guys when you do 4 C2 you get number of sides also minus number of side will give you number of triangle now they are saying number of Triangles diagonals are 44. You tell me number of sides. So guys, I don't know. But if there are n sides to draw a diagonal, I need any two corner points. N sides means n corner points. If you have n corner points, you need any two corner points. This will not give you diagonals. Along with the diagonal, this will also give you n sides minus n is equals to 11. Sorry, is equals to 44.

[00:31:52]Now I will go from option to question.

[00:31:54]The option which will give me 44. that option will be correct. One of the option is 11. Let's put 11 over here. 11 C2 guys. 11 C2 11 into 10 divided by 2.

[00:32:06]11 C2 is 55 minus in the place of N you will put 11 uh 55 minus 11. Are you getting as 44?

[00:32:16]Yes. So since this option give you LHS equals to RHS this option is correct.

[00:32:23]Rest all the options are wrong guys.

[00:32:25]Even you use one shortcut guys write down number of diagonals is equals to guys n into bracket n into bracket n - 3 upon n upon 2 guys you can use this guys if by any chance this things do not trigger you are unable to get this guys and if they're asking number of diagnos just put the sides you will Get the diagonals am this time they have given us the diagonals are 44 do you tell us side in this case also we will go from option to question but guys this is very easy to put n into nus 3 upon you you can use this formula also guys if you put 11 guys let's put guys let's put 11 here let's put 11 guys 11 - 3 everybody guys 11 - 3 ult* 11 divided by 2 since this option give you LHS equals to RHS. So guys this option is correct rest all the options are wrong everybody guys in this way you can solve this you can solve this guys is this point clear to all of you can I move forward the number of diagonals having six sides am sir we can just put the six over here and you can get the answer sir six sides means six corner point to draw one diagonal we need two corner points in how many ways you can pick two corner points out of 62 into minus number of sides how much you will get 6 into 6 into 5 / 15 - 6 I'm answer we will get nine diagonals yes 9 is the right answer even you can use the formula number of diagonals is equals to n into bracket n - 3 upon 2 let's put 6 6 - 3 3 divided by 2 that's 9 and here also 9 whatever you feel comfortable you can apply I Yes guys there are the number of triangles that can be formed by choosing the vertex.

[00:34:39]There are 12 points out of which seven are colinear to draw the triangle we need three points but at least one should be the colinear. I have given you one crux in the class. You don't get the triangle when the three points are taken from the column. Rest all the points give you triangle. Ammon sir, how to deal with this? If you do like this total ways minus ways which don't give us which don't give us triangle. Amen sir.

[00:35:16]Aman sir to draw the triangle what we require? We require three points out of 12 points. But all the three points will not give you triangle. If these three points are taken from that seven points, the triangle is not formed. Minus when the three points are taken from that seven points, we don't get the triangle minus 7 C3. Guys, what is 12 C3? 12 into 11 into 10 divided by 6 220 - 7 C3 7 into 6 into 5ID by 6 I'm answer - 35 how much 220 - 35 you will get guys 185 ways 185 base everybody guys is this point clear to all of you 185 ways here is the answer guys is this point clear to all of Is this point clear to all of you guys?

[00:36:11]Can I move forward? Guys, can I move forward? The number of parallelograms that can be formed from six parallel line. A man sir, we are having six parallel lines. 1 2 3 4 5 6 intersecting another four parallel lines. 1 2 3 4.

[00:36:35]What is our requirement to draw parallelogram? We need two parallel lines guys. We need two parallel line on x-axis and two parallel line on yaxis.

[00:36:52]Guys, how many lines you are having here? Six. Out of six, how many ways you can select any two among 6 C2. Wait, till now paradog is not formed. We need and and means multiply or you can say we need simultaneously we need two more lines in the y-axis. How many lines are there in the yaxis? Four. Out of that you need any two 4 C2 guys. How much you will get? 6 C2 answer 6 into 5 / 2.

[00:37:19]Answer 15 * 4 into 3 divided by 2 you will get 6 15 into 6 I'm answer we are getting 90 ways everybody guys we are getting 90 ways and guys here we are done with the problems based on geometry guys successfully we have completed problem based on word number geometry guys can I move forward everybody with me guys can I move forward forward guys. Can I move forward? Are you all here with me? Guys, can I move forward with everybody? Guys, can I move forward?

[00:37:59]Are you all here?

[00:38:03]Yes. Amir, let's start with the problems based on circular permutations. Now guys, with me, in how many ways, guys?

[00:38:13]In how many ways uh can a party of four men and four women be seated in a round table so that no two women's are together. I have already told you whenever you are solving the questions on circular and question has given you restriction. Plot the kind which is having the restriction. Ammon sir here women's are having the restriction. So plot the woman everybody draw a circle.

[00:38:39]Plot the woman am so that no two women's are together no two women sorry plot the quantity which is having no restriction here women's are having restriction don't pl plot the women's men's are free how many men's four men's m1 m2 m3 m4 in how many ways four men's can form a circle amount four men's can form a circle 4 - 1 factorial done with the formation of the circle now guys no Two women's are adjacent. It means there should be at least one man between one woman. So guys, how many places are there for the woman? Let's count. One 2 3 4 guys in how many ways four womens can sit in the four places? Four factorial. Ammon should I do minus one again? No. Why? Because you the woman have to sit in the same circle and circle is already formed. If I have to make them sit in a new circle then I will do minus one. Guys minus1 is done to form a circle and you have already done that. If you do minus1 again you will form a new circle and you have to make the sit woman in the same circle.

[00:39:50]So guys in how many ways four women can sit in four places? 4 P4 sir 3 factorial ult* 4 factorial 24 * 6 everybody guys 144 ways guys 144 ways option C is the right answer everybody is this point clear to all of you guys is this point clear to all of you can I move forward yes I'm answer now the number of ways five boys and five girls be seated in a round table so that no two Boys are adjacent.

[00:40:23]This time you are having restriction in boys. So plot the girls first. Plot the girls in a circle. Form a circle guys.

[00:40:30]Plot the girls. Sir G1, G2, G3, G4, G5 and G6. In how many ways?

[00:40:41]Six girls can form a circle. Sir 6 - 1 factorial.

[00:40:47]Now how many places for the boys? 1 2 3 4 uh sorry by mistake I made six guys five uh so that no two boys five girls 1 2 3 4 four and five guys in how many ways five girls can form a circle sir five minus one factorial now how many places for the boys let's Count s they should not come together. It means someone should be there. At least one girl should be there. 1 2 3 4 5 is the five boys occupy this places. They will never come together. So guys what is five fac five in how many? Five boys can sit in a five places. Five p five that is again five factorial. So guys four factorial multiply by 5 factorial guys 24 * 5 factorial 120. Everybody guys 2880 Everybody guys 2880 guys look at this part first past look at this part very very much important part guys look at this guys now now let's start with the problems based on theorems guys everybody here now I have drafted us so many questions in this in this we'll cover all the things in this we'll cover alternate non-alternate Right guys, one particular thing occur never occur different objects grouping distribution everything will be covered here. Now let's start with this everybody. In how many a man has five friends in how many ways he may invite one or more? Guys, I have given you one shortcut. In how many ways he may invite one or more? 2 raised to the power n minus one among s out of five friend he may invite anyone. 5 C1 plus two friends 5 C2 5 C3 plus 5 C4 plus 5 C5. In this is in this way he may invite one or more than one friends. But if they would have given zero or more, I would have started from 5 C 0 + 5 C1.

[00:42:59]Sir, for this only you have given us one shortcut 2 raised to the power n minus one. So what is 2 to the power 5 - 1? 32 - 1. Everybody guys 31 is the right answer. Everybody 31 is the right answer. Guys, can I move forward everybody? Yes. Now in an examination there are guys 10 questions consisting of in this six questions are of the algebra and guys and and four questions are of the geometry.

[00:43:34]If in uh at least one question guys, at least one question guys, this word and one or more both are same. At least one or one or more than one guys, both have the same meaning that a student has to perform at least one question from each section. At least one question from each section. At least one means one question, two question or all the question guys. At least one or one or more than one both have the same interpretation is to be attempted guys.

[00:44:06]It means a person has to attempt at least one question from algebra and at least one not or if there is word or it comes in my mind let's then I will say add here I will say 2 raised to the power n minus one at least one and at least one from the geometry 2 raised to the power n minus one he has to attempt at least one question from the algebra and geometry if they would have said any one section either of geometry or algebra Then I would have added either he attempt any one question of algebra or he attempt any one question from the geometry then I would have added he has to attempt at least one question from both this and this both guys that's why here there's word and so I will multiply I'm answer two how many questions 6 - 1 multiply by how many questions are there here 2 to the power 4 - 1 what is 2 to the^ 64 - 1 everybody guys 63 2 raised to the^ 16 - 1 guys that is 15 63 into 15 everybody 945 is the correct answer. Yes sir. Same as 20. Yes. Same same same same question as yes everybody. Next question. In an examination candidate has to pass each of the four paper. Same situation that you have you have to pass all the papers. four papers in how many different ways he can fail. Ammon in how many even if he fail in one subject he failed. If he failed in two subject he failed. If he fail in all the three subjects he failed. If he fail in all the four subject then also he can fail.

[00:45:48]In how many ways he can fail? He can fail in four ways. Either out of four he fail in one out of four he fail in two.

[00:45:55]Out of four he fail in three. Out of four he fail in four. From 1 to n it is.

[00:46:00]Do you have any shortcut? Yes answer.

[00:46:02]Again 2 to the power n minus 1. So that you have 2 ra to the power 4 - 1 everybody 15 ways.

[00:46:12]Is this point clear to all of you? 15 ways is the right answer guys. Yes. Yes.

[00:46:17]Yes. Taresh will see that also now guys a very beautiful question. Mark this question as an LDR guys. Not this one I think. Okay. Now let's see guys let's look at this question.

[00:46:33]Just a second. Uh 26. Yes guys a very good question. Now mark this question as an LDR. Mark this question as an LDR.

[00:46:43]Guys look at this question. In an election the number of candidates candidates are those who participates in the election. One or more than one number of members. Members are those who get elected. Candidates are those who participate in the election. If voter has guys, if voter has 254, if voter has 254 different ways, the number of candidate, let us say guys, I don't know how many candidates are there. Let us say there are n candidates. Okay. Now what they are saying u one or one more guys candidate is one more than the number of members. For example, if four are going to become the member, candidate is one more, candidate are five. If number of persons who are going to be elected is six, candidates are seven, candidate is one more. So guys, in how many ways a voter can vote? They are saying find the number of candidate. If the voter can vote in 254 ways, a voter can vote maximum how many ways? The number of persons who are going to get elected.

[00:47:52]If for example guys for example I have to go from option to question sir let's say if I say option A if I go from option A let us say option A if I assume that option A is correct if I assume option A is correct guys how many ways a voter can vote among a maximum voter can vote the maximum uh if there are eight members eight members or eight candidates how many members will be there seven so a vote Voter can vote maximum seven votes. In short, if I will go a voter can vote any one vote out of it, he can vote any two vote. He can give any three vote. He can give any four vote. He can give any five vote. He can give any 6 vote or he can give any 7 vote. But guys, he don't have the option of 8 C8. Ammon the shortcut is 2 to the power n only when here you have 8 C8.

[00:48:51]This time you don't have 8C yet. What?

[00:48:53]Because a voter cannot vote all the candidates. Voter can maximum vote the number of persons who are going to become member. Member is going to be one less than the candidate. So voter can vote maximum seven. This time you don't have this 88. What is the outcome of 88?

[00:49:09]Outcome of 838 is one. So guys can I say this will come here and here guys voter is having this much number of ways. Guys this is now shortcut. A voter is having 2 to the power n minus 2 and this is guys 2 raised to the power this is 254.

[00:49:24]Now I will go from option to question the option which will give me 254 that will be correct. I answer Y minus Y minus 2. This time voter is not having two ways. He can't 8 C 0 plus 8 C8. 8 C8 is also one and 8 C 0 is also one. If voter cannot zero vote and voter cannot give all the votes. So guys I have subtracted two. Now I will go from option to question guys. 2 to the power 8 - 1. Are you getting 254? 2 to the^ 8 guys. 2 raised to the power 8. 256 - 2 254. Yes, this option give me 254. So this option is correct. Rest all the options are wrong. Yes guys a very beautiful question. Even if guys this main part is this one. You have to form this one guys. This time voter is if voter is having all if voter can vote all the candidates he is having how many ways 2 raised to the power n minus one but this time he is not having one more way he can't wait he can't vote all the candidates he can't vote all the candidates that is 88 so minus one again more minus one so guys minus2 everybody guys is this point clear a very beautiful question guys mark this question as nlddr raised to the power Five LDR raised to the power five.

[00:50:52]Sir, how we will be able to say that a voter can vote for all the candidates.

[00:50:58]Sorry, a voter cannot because see see how can you vote? Common sense. If there are going to be eight out of eight, if the seven persons are going to become member, if voter can give all the eight votes, then how the members will be seven? If I if you are going for any election, okay, there are five parties.

[00:51:19]Now how and guys there are total eight candidates but you can vote only four because only four will become member. If all the common people will give all the eight votes then everyone will become a member. They are clearly saying the members will be one less. If so you have to select out of five you have to pick the four best one not all the five. Then there will no vote section. If everyone will vote for everyone then how will we will pick the best four common sense?

[00:51:49]Yes. LDR means last day revision. Again you have to do it twice one day before exam. Now guys next question. A boy has three guys. Lamar this question also has an LDR last day revision twice. Once after this class one day before exam also guys see one day before exam you have to solve entire practice batch question again. That will be very good option. Now understood sir. Understood sir. Okay.

[00:52:14]Now everybody guys with me. Everybody guys with me. Now a boy has three library tickets. Okay. A boy has three library tickets and eight books of his interest in a library. He does not wants to borrow mathematics part two unless he has borrowed part one. So what is the condition? If you want to borrow mathematics part two, you have make sure that you have borrowed part one. Okay.

[00:52:42]In how many ways ch he may choose three books guys out of it he has to select any three. He has three library tickets.

[00:52:50]Now see guys there can be three scenarios.

[00:52:54]What if I say I want maths one. I want maths one but I don't want maths two.

[00:53:01]Guys, it is not compulsory that it is not compulsory that you have to borrow maths 2. But yes, I can say I want to borrow maths one and I want to also borrow maths 2. Amara, can we draw the case? I just want to borrow maths 2. No, because if you want to borrow maths 2, maths one is compulsory. Yes, you can borrow maths one but you don't buy maths 2. Maths one has no compulsion that you have to buy maths 2. But if you want to buy match two, match one is compulsory.

[00:53:30]Third case, I don't want match one also.

[00:53:34]I don't want match two also.

[00:53:39]10 minutes.

[00:53:41]>> Okay guys, I don't want match one and even I don't want max two also guys. Now here out of eight you need to select three. Out of eight you need to select three. What is the max one? I want one book one person will compulsory come. I will say out of it max one go to the ground guys ground one particular thing always occur and one particular thing never occur. Max one come in my bag. Match two go out. If out of eight you have already selected one. One is you have thrown out. Match two thrown out. How many total are left?

[00:54:20]Six. But this time you need to select only maths one is already selected.

[00:54:25]Maths one one particular book is already select. Math one is already there.

[00:54:30]You need to select three. But maths one is already there. And you have thrown maths two out. I don't want maths two.

[00:54:36]Now out of six I need to select only two. Why? Maths one is already there. 6 C2.

[00:54:45]Sir I can't understand because Tarish if you have not studied the logic number nine and 10 sorry this question is out of scope one particular person occur never occur guys again I'm saying without studying from me or without studying from the YouTube revision if you join this pack this is waste for you because I am going to use that terminologies which I have taught guys everybody again out of eight I need to select the Three.

[00:55:16]First scenario am they have said if you want to borrow maths 2 then maths one is compulsory then it is mandatory that you have to make sure that you have borrowed maths one. Now there can be three scenario. I want maths one but I don't want maths two s I want match two for which match one is compulsory. I don't want any. Now there is no more case no other case. Most beautifully drafted question from the logic number nine and 10. one particular book occur and never occur case guys. Now next question answer I want maths one also and I want maths 2 also out of eight I will say librarian give me maths one maths two already aside you want to select three but two books are already selected now you need to select one more out of when you already selected maths one and two out of eight now you two are already in your bag how many are there six out of six I just want to select anyone I don't want M1, M2. I as a librarian throw this match one and match 2 from my in front of my eyes. Don't show them.

[00:56:24]Now, how many books are left? Six. Till now, I have not selected any plus 6 C3.

[00:56:32]Everybody with me? Let's solve this.

[00:56:35]What is 6 C2 guys? 6 into 5 / 2 answer 15.

[00:56:42]I'm answer 15 + 6 + 6 C3 6 into 5 into 4 / 6 20 how much guys 20 + 15 + 6 everybody 41 ways guys 41 ways are you all here with me? Yes. Yes.

[00:57:06]Yes guys, a very beautifully drafted question. Mandatory you have to do it.

[00:57:12]Mandatory you have to do it guys. Is this point clear to all of you? Is this point clear to all of you?

[00:57:21]Yes. Yes. Yes.

[00:57:25]Guys, everybody with me? Is this question clear to all of you? Can I move forward?

[00:57:31]Yes. The answer 5 C1 + 5 C2 + 5 C3 + 5 C4 + 5 C5 everybody okay Tarkesh now 5 C1 + 5 C2 + 5 C3 + 5 C4 I'm answer here this is just a shortcut 2 to the power n - 1 guys how many 2 to the power 5 - 1 32 - 1 everybody 31 ways 31 ways everybody option B is the right answer 31 ways is this point Clear guys?

[00:58:01]Is this point clear? Now guys, make a one correction over here. In how many ways? Five price. Guys, make a correction. Five price. If you have here three, convert this into five. In how many ways? Five prizes can be distributed among three persons equally.

[00:58:18]A sir, if you want to distribute five prizes among three students, I'm sir, how can we distribute equally? One one is only possibility. If you any student will get two prize two prize one will get only one price. If you want to distribute equally two prize will be always left over. Out of five you need to select any three and you have to distribute any three. Out of five I need to select three. Now I have to distribute among three person. Order is important. Which prize will go to which person. So out of five I just need to select three and arrange three. 5 P3 plus 5 into 4 into 3. Are you all getting 60 ways?

[00:58:58]Guys, are you all getting 60 ways?

[00:59:02]Yes. Five prizes can be equally divided among three students.

[00:59:07]Yes. Now, next question guys. Three girls. Everybody with me? Amen. Sir, three girls, five boys are to be seated in a room so that no two girls sit together. Guys, can you guys there are five girls and five boys? Sorry, there are guys three girls and five boys to be seated in a row so that no two girls guys. Can you tell me is this a question of alternate or non-alternate?

[00:59:44]Guys, is this a question of alternate or non-alternate? Guys, this is a question of alternate or non-alternate.

[00:59:52]Everybody guys with me? This is a question of alternate or non-alternate.

[00:59:58]Ammon sir, there are two kinds boys and girls and I have not drafted any condition on boys only condition on girls. If there is a restriction on only one kind this is nonalternate it means no two girls are together boys can come together. Guys, this is a question of non-alternate.

[01:00:21]Yes, non-alternate. Plot the quantity.

[01:00:23]What is our method? Plot the quantity which is more in number. Amaz boys are more in number. B1, B2, B3, B4, B5. In how many ways? Five boys can be arranged in five places.

[01:00:38]Five P5.

[01:00:40]Now let's count the places for the girls. Since it is nonalternate, consider all the corner places. 1 2 3 4 5 and six. In how many ways? Three girls can be arranged in the six places. * 6 P 3 5 5 P5 5 factorial that is 120 * 6 into 5 into 4 120 1440 guys is this point clear everybody yes not 720 Turkish 120 into 120 5 factorial 5 P5 Now moving forward Now in how many ways guys? Now this four boys and three girls can be sitted in a photograph. Alternate. Now they have clearly used the word alternate. Now guys in alternate I have taught you two things. Same quantity or different quantity. Among four boys, three girls.

[01:01:43]Four boys, three girls. They are having different quantity. So guys you have to plot the quantity which is more in number. Four boys. B1, B2, B3, B4. In how many ways? Four boys can be arranged in the four places. Four, P4. Since they are asking alternate and you are having different quantity. Don't consider any corner place. Just take this places. Don't consider this corner places guys. Since they are different in number sir three places for the three gigles multiply by 3 P3 4 P4 4 factorial 3 P3 3 factorial 4 factorial 24 * 3 factorial is 6 everybody guys 144 ways everybody guys 144 ways 144 ways everybody is this question clear to all of you can I move forward Guys, can I move forward with me?

[01:02:50]Everybody with me? Can I move forward?

[01:02:52]Yes or no? Yes. A sir. Now, yes. In how many ways? 12 prizes can be distributed among four students. So, they have uh each have three prizes guys. I have to distribute. This is a question of the grouping. Then you have to distribute also. How to solve the questions of the grouping and distribution guys. n factorial upon p factorial into q factorial into r factorial. Here you will have extra m factorial if you have to if you have made any equal groups. Now you have to distribute group this groups into n persons multiply by guys n factorial guys something like this r guys guys like z factorial anything it can be answer how many things you have to distribute 12 factorial everybody with me divided by all the persons four students will get three three prize three factorial into 3 factorial into 3 factorial into 3 factorial guys you have made How many equal groups? Ammon sir, we have made four equal groups. Multiply by four factor. Now you have four different different packages. Now you have made four groups. Now you have to distribute four groups into four person. In how many ways you can distribute four groups in four person? Multiply by 4 P4. 4 P4 is 4 factorial. This will cancel. Now let's solve this in the calculator. 12 factorial 12 into 11 into 10 into 9 into 8 into 7 into 6 into 5 into 4 into 3 into 2 into 1 divid by 3 factorial 3 factorial 3 factorial 3 factorial means 6 divided by 6 divid by 6 divid by 6 divid by 6 are you all getting everybody are you all getting 3 6 9 600 36 9600 guys 36 96 600. Is this point clear to all of you guys? Can I move forward?

[01:04:55]Everybody, can I move forward? Yes. Now again, again a question of guys. This was a question of grouping plus a distribution. Again a question of grouping and distribution. Guys, this was a question of fixed distribution.

[01:05:16]Now this is a question of not a fixed not a fixed distribution. In this question I have mentioned that each students will get get three three prices. This is a fixed question has told you that every will get every much will every person will get three three three object. Now in this question guys this is a tricky area and guys this is a question that they ask out of the box question in the year MTP 2024 guys. MTP mock test paper 2024 everybody guys shall we start now five balls five balls of different color are to place in three boxes guys five balls it just like you have to distribute five balls among three boxes or you can say three persons okay of different size each box can hold all the five balls each box can hold all the five balls And no box can remain empty. In how many ways you can do this grouping? In how many ways you can do this distribution?

[01:06:22]But they are silent. How which box can hold how many? They have clearly said no box can remain empty and each box is having capacity to hold all the five balls. A sir what is the difference between this question and this question?

[01:06:37]This question has said each student will get three three prices. And here we have to think how many cases are possible. So guys here let's start everybody with me I'm answer sir let's say each can hold any let's say first box will hold one ball second will hold one ball out of five you have given one one sir why you are not given zero because no box can remain empty then in that case third have to hold three balls now what if you say this will have one ball this will have two ball you are having total how many five balls two and 1 three. Now you're left with only two balls. Any other case answer? Let's say this will have zero balls. No, this will have one ball. This is four. No, this is not possible. Question has clearly said the box cannot remain empty. If the box can remain empty, I would have directly applied the shortcut RA to the power n.

[01:07:34]But since the box cannot remain empty, I have to draw this cases. I have to think which box can hold how many box as per the situation given in the question.

[01:07:45]Guys, what is the difference here?

[01:07:46]Question has mentioned that each student will get three. Three, three, three, three. And here I have to think how many how many? How many? Which box will get how many? I have to think. This is not a fixed. I have to fix it. And here it was fixed. I was not worried about to fix.

[01:08:03]This is the basic difference. Now let's do it. Am how many balls are there?

[01:08:09]Five balls. First box will hold one ball. Second box will hold one ball.

[01:08:15]Third box will hold three. Third three balls. How many equal groups we have made? Amaz by mistake we have made two equal groups. Multiply by two factorial.

[01:08:24]Again guys multiply by. Now these are the three packages. Now you have to make this. You have to distribute these three packages among three boxes or I can say three students. Three box in three students. 3 P3 that is three factorial.

[01:08:39]Now we have to distribute this five balls.

[01:08:45]First person will get one ball. Second will get two ball. Third will get two balls. How many equal groups am we? We have two equal groups. Multiply by two factorial. Now you have to distribute these three boxes among three persons.

[01:08:58]Multiply by 3 factorial. Guys, let's do it. 5 factorial 120 divided by 3 factor that is divided by 6 divided by 2 * 3 factorial is mult* 6 I'm answer the outcome of this is 60 now 120 divided by 1 divided by 2 divided by 2 divided by 2 * 3 factor that is mult* 6 are you all getting 90 yes am sir so how much you will get 90 + 60 we will get 150 ways is I'm answer most challenging question of the MTP is till now many students don't study this but you know my nature I teach whether it is having possibility of 1% of being asked we'll discuss in the class guys I hope with this variations we are covering all the theorems particular thing occur never occur guys uh 2 to the^ n minus one everything everything grouping distribution alternate non-alternate Everything ammon sir let's solve this one guys in how many ways five four boys and three girls let's change this guys let's change this what if you have four boys and four girls guys what if you have let's let's change the question what if you have same quality four boys because this same question we have done here here guys the data was same and they have asked alternatively here the data was different sorry but what if you have the same data you have four boys four And again I'm asking alternate guys plot the quantity which is more in number among sir when you have the alternate they are ask alternate and you have the same quantity both the kinds are having same quantity there's a tricky area now let's plot any one kind let's plot the boys guys now let's see this question guys in how many ways four boys and four girls plot any one kind plot the quantity which is more in number but both are same in number so plot four boys B1 B2 B2, B3 and B4 in how many ways guys?

[01:11:03]Four boys can be seated in a four places. A man sir four P4.

[01:11:10]Now you need to arrange four girls. But since I am asking alternate let's count the place. 1 2 3 I only stopped you to consider the corner place. But there are four girls and we are having only three place. So you have to consider any one corner place guys. Consider only any one corner place. Don't consider both the corner place. Consider any one corner place. Now in how many ways four girls can be seated in a four places? 4 P4.

[01:11:41]Now I'm answer later on multiply your answer by two. Why two? See if the girls occupy the this if you consider last corner place the alternative is boy girl boy girl boy girl boy girl okay but if you consider this not if you consider first corner place is girl boy girl boy girl boy both are alternate so guys whenever you have same quantity consider any one corner place and multiply your answer by two and guys if you have different quantity don't consider any corner place they will never ask the difference of more than one whenever they are asking alternate. If they have given a difference of more than one like they have given here difference of two it will be always a question of non-alternate. So guys this is your answer. 4 P4 is 4 factorial 24 * 24 * 2 everybody you will get the answer 1 1 52 1 52 everybody option A is the right answer 1 1 52 yes is this point clear everybody now let's have a small break we'll connect after break case guys let's move forward everybody guys before break we have concluded all the variations of the problem now guys let's start with the discussions of the miscellaneous problems on permutations and combinations everybody guys are you all here can I move forward guys can I get The yes yes yes yes guys all the backups will be uploaded in the back back uh in your app before five or six this backup will be uploaded I promise okay now let's move forward am sir now guys this questions are going to be technical please pay attention how many sixdigit telephone number can be formed using 10 different digits guys whenever the question is silent we always assume in case of four number the repetition is allowed you have to form a six 1 2 3 4 5 6 digit four number Aman sir even the zero can come in the beginning in case of four number it can start with zero also we have solved this kind of question in the class any number can come in the beginning how many different numbers you have 10 10 numbers can come 10 numbers can come 10 numbers can come 10 numbers can come 10 numbers can come 10 numbers can come in short 10 raised to the power 6 must be the answer. But the problem is guys they have given 10 raised to the power 8 they have given 6 raised to the power 10 answer must be 10 raised to the^ 6 they have given 10 C9 that's no sense amir here we have assumed repetition is allowed repetition is allowed what if repetition is not allowed all the numbers should be different guys in that case I just need to pick out of 10 I need six numbers and I have to arrange them so guys in that case the answer will be repetition is not allowed 10 P6 and the answer will be correct. If they would have given both the options, I would have preferred this one because in the case of phone number, we should be always assumed that a repetition is allowed. And even in this question guys, they have not stopped us. And guys, even in this question, they have not stopped us. Amen. Sir, what if we draw like this? What if we solve like this? 1 2 3 4 5 6 How many numbers can come? 10 the number which has already taken place cannot come now. 9 then guys 8 then 7 then 6 then 5 guys both are same either you solve this or you solve this guys both are guys both are same guys both are same. So guys since option D matches I will say option D. But yes, the better answer for this must be a because in the case of password and in the case of phone number if the question is silent we will always assume that repetition is allowed. Now no no no no no evening session guys.

[01:16:02]Today we have to conclude this. Okay.

[01:16:03]Now in a lawn uh in a lawn in a lawn different ways four persons can stand in a line. Guys, in short, I just need to select and arrange four persons in four places. Guys, four P4 that is four factorial. Guys, 24 is the right answer. Everybody guys, 24 is the right answer. Now, company simult to top simultaneously promote three of its eight department assistant manager. Guys, there are eight departments. There are eight departments and company has to promote their uh assistant managers of any three. Company wants to promote the assistant of any three. In how many ways the promotion can take place? Amen. The question is silent. Should I arrange? I will say give out of eight I just need to select three. 8 C 3 guys. What is 8 into 7 into 6 divided by 6? I'm answer we are getting 56. Option B is the right answer. But by mistake if you say 8 P3 guys you will get 336 now. So that is also in the option guys. Now I have already said guys see now you have to apply your brain. This can never out of eight departments guys. You need to promote any three order is not important. Either you promote A, you promote B or you promote C or you promote C first then D then A. Promotion remains the promotion of the persons remains the same. Guys in this I will say order is not important either you first put A B C B A C CA C CA A B promotion of the persons are same I just need to guys have how many guys the three different promotions can be made sequence of the promotion is not relevant so guys in that's why here option B will be the correct answer yes yes yes option B will be the correct answer guys a very good question I will suggest make sure that you revise this first LDR Next question. A committee of eight from a committee of eight persons. Form a committee guys. Form a guys. It should be form if form a committee of eight persons in how many guys? There are we have to form a committee of eight persons. From eight persons we choose a chairman. We need one chairman and a vice chairman.

[01:18:32]Assuming that one person cannot hold more than one position. If a person has become chairman, he cannot become vice chairman. If a person has become vice chairman, he cannot become a chairman.

[01:18:44]Guys, it means now I need to make a team of eight from it in which one should be chairman, one should be vice chairman and others are member. So guys, in how many ways you can select one chairman out of it? Out of it, you can come, you can come, you can come. Out of it, I want one. 8 C 1 1 2 3 4 5 6 and seven.

[01:19:12]By chance guys this person has become chairman. Now one person can hold hold only one position. It means repetition is not allowed. The chairman now cannot become a vice chairman. Now how many persons are available to for the selection of chairman Aman? for vice chairman. Now you have seven out of seven you need to select anyone seven for example guys any one person has become a vice chairman. Now how this how many ways six members can be selected out of six members multiply by guys 6 C 6 8 into 7 into 1 everybody guys you will get 8 into 7 into 1 guys everybody 56 is the right answer everybody guys 56 is the right answer I request all of you mark this question also as an LDR question yes mark this question also as an LDR question moving forward everybody Guys, moving forward guys. In how many ways? Four letters are written. Four letters are written and four envelopes are addressed. In short guys, I have to distribute four letters in four envelopes. Okay. The number of ways in which all the letters do not go in the correct envelope. Guys, there is total ways in how many ways? Four letters, four envelopes can be arranged in four places. Guys, consider this envelopes as a place. In how many ways?

[01:20:43]Four letters total. In how many ways?

[01:20:45]Four letters can be arranged in four envelopes. 4 P4 4 P4.

[01:20:52]I'm answer guys wait 4 P4 24 is the total ways. But there is only one way when the light right letters will go in the right envelopes. have only one way that the A letter goes to A address, B address goes only one way only one way is the correct way. So guys, how many incorrect ways are there?

[01:21:12]23 are the in they are do not go in the correct guys. 24 are the total ways in one way only they can go in the correct envelope that yes right let us go into a right envelope guys. So how many incorrect ways are there? 23 ways. Now this is the best way to solve this question. But many students say ammon sir can we solve this question in a form of uh grouping form of group then distribute. Yes we can do it like that also am sir we need to distribute four letters among four persons or we can say four nvers so that each will get one one factorial 1 factorial 1 factorial into 1 factorial. How many equal groups? Four equal groups.

[01:21:56]Now you have four groups. Now you just need to distribute this four groups among four person 4 P4 again multiply by four factorial so that this will cancel and at the end you will get 24 only these are the total ways one is the correct way. So how many incorrect ways are there guys? 23 are the incorrect ways guys. Many students sometimes say among can we consider this is a question of grouping and yes but the point is guys since all the envelopes will have 11 one only. So even if you write like this both are same everybody moving forward the number of ways in which eight examination paper be arranged so that best and wor best and worst paper never come together am sir we have to do total ways minus ways when they come together when they come together in How many total ways you can arrange eight eight papers in eight places? 8 P eight that is 8 factorial. If best and worst come together. Now you put them in one box.

[01:23:08]Now how many are left out of eight? You put the two in box. Six are left. 1 2 3 4 5 6. Let's count guys. Let's count. 1 2 3 4 5 6 7 ammons are seven factorial.

[01:23:25]Even if they arrange inside the box they are together multiply by 2 factorial. So the correct answer 2 factorial outcome is two 7 factorial into 2. Guys can you see option A? 8 factorial - 7 factorial * 2 or 2 * 7 factorial both are same everybody option A is the correct answer. Yes guys no no no not B not B.

[01:23:47]Not B guys. No no no Tesh not B. Even if they are apply even if they are inside if they inside are in the box still best and worst are together. So guys multiply by two factor guys don't make this mistake. Now guys next question a box contains guys a box contains seven red, six white and four blue balls. How many ways selection of three balls of each color guys? You have to select three balls of each color. It means guys one should be red and one should be white and one should be black. So guys I will say one should be red 7 C1 multi one should be black 6 C1 4 C1 guys we are saying that they are of the different colors are same but the size all balls are different of of different size. Okay, let this thing 7 into 6 into 4 everybody guys 7 into 6 into 4 everybody guys 168 is the right answer 168 guys 168 option D is the right answer I hope this question is clear to all of you guys can I move forward yes guys 12 school teams are participating in a quiz contest the number of ways first second and third position may He won that's out of 12 in how many ways of you can first 12 participants are there and I have to select and arrange also am this time question is silent now should I apply the arrangement I will say no first let's do the selection 12 C3 now guys what is 12 C3 am 12 into 11 into 10 divided by 6 you are getting 220 now option don't match let's apply the arrangement let's arrange three persons also. So guys multiply by 6 here you will get 1320.

[01:25:49]Aman sir what if Aman s what if they have given 220 also. Now apply your common sense guys. See now 12 school teams are participating in a quiz first second third.

[01:26:01]For example school A B and C are selected. Now guys it's not if the know the sequence will matter. If you place A here A is first B is second C is third.

[01:26:11]If you the same persons are selected but if you place A here B here C here this is a different picture one in one picture A is first in second picture here order is important. Now guys in something I can say after selection I have to distribute these things among a person the order is important. So guys I will say 12 C3 or you can directly say out of 12 I need to select and arrange three prices 12 P3 the answer will be 1 320 guys both are same. Yes everybody is this point clear to all of you? Guys is this question clear to all of you?

[01:26:52]Yes yes yes yes guys. Moving forward next question. A box contain guys moving forward the next question.

[01:27:02]Okay, we are done with this. Yes guys, many questions are waiting for you. So many questions are waiting. I have drafted almost guys. I have drafted 30 questions additional just to cover the miscellaneous, miscellaneous and miscellaneous. Let's do it guys. A box contain guys there is one box contains three pink, two purple and four orange caps. In how many ways this can be arranged so that caps of same color come together that's in their distinct okay color same are same same color so I'll put three pink in one box two purple in one box and as four orange in one box since you want them together count that box as in one unit 1 2 3 am three factorial even if they are arranged inside the box they are together if by mistake they would have not said they are of the they are not identical If they would have said they are identicals, I would have not arranged anything inside the box. The three factorial will be only right answer. But since they are saying they are not identicals. Yes, since they are not identicals, I have to apply the arrangement. Even if they swap inside the box, they are together.

[01:28:15]2 factorial multiply by 4 factorial guys. 6 into 6 into 2 into 24. Everybody 728 728 Everybody option B is the right answer. Yes guys option B is the right answer. Moving forward the 12 questions are to be answered in yes and no. In how many ways this can be answered? Let's check it down. How many questions are there? 1 2 3 4 5 6 7 8 9 10 11 12. Guys, just a second.

[01:28:51]1 2 3 4 5 6 and so on. Guys, there are 12 questions guys. Each question can be answered in yes or no. Either you can say yes or you can so. So guys, question number one can be arranged in how many ways guys? Sorry. Question number one can be answered in two ways. Second question can be also answered in two ways. Two ways and so on. I answer all the 12 questions can be answered in two to two ways. So guys, how is the total 2 raised to the power 12?

[01:29:21]What is 2 to the^ 12? 2 into 1 1 2 3 4 5 6 7 8 9 10 11 12. Are you all getting 4096? Everybody guys right answer 4096.

[01:29:32]Balam right answer 4096. I'm answer but what if I don't want to draw the places?

[01:29:37]See in how many ways? See out of 12 questions in how many ways you can select any one question among 12 C one guys. No no you have to do it like this only. You have to do like this only. The rest it will be very lengthy. Yes. Now the number of ways guys six boys four girls be arranged in a row so that all the four girls are together. Put the girls in one box. G1, G2, G3, G4. So all the girls are together. How many boys are left? B1, B2, B3, B4, B5, and B6.

[01:30:17]Let's count. Six boys and this one is seven factorial everybody.

[01:30:23]Even if they arrange inside the box, still girls are together. Multiply by four factorial. Question was that all the four girls are together. So guys 7 factorial into 4 factorial everybody option C is the right answer.

[01:30:37]Moving forward in how many ways in how many ways three peoples can form a uh three peoples can be formed from a group of five people. Guys what they're saying in how many different groups can be formed guys you have to form a group. Okay. Group of three peoples can be formed out of five people. I just need three person out of five. In how many I can survey? I can select any three person. Five C 3 guys.

[01:31:08]What is 5 C3? 5 into 4 into 3 divided by 6 10 why not 5 P3? Guys I just need to form a group A B CDE E three person groups I have to form guys. That's all.

[01:31:20]So guys in how many ways? A group of three. I just need to select three groups. Three persons. Either ABC come or B A C come or CAB come order is not relevant group is same. So guys here arrangement is not required and even if you multiply by three factorial that is 60 60 is not in the option guys everybody option C is the right answer.

[01:31:40]Now in how many ways? Four people, four people can be selected at random from six boys, four girls.

[01:31:53]In how many ways? As six boys uh four people I have to select four people in how many ways? Four people can be selected at random from six boys and four girls. If there are exactly two girls in short in this form the two are boys and two are girls. In how many ways you can select two boys out of six? 6 C2 two girls out of four 4 C2 6 into 5 / 2 15 multiply 4 C2 4 into 3ide by 2 guys 6 15 into 6 guys. Are you all getting 90 ways? Yes. Are you all getting guys 90 ways? Yes, Aman sir. 90 ways everybody.

[01:32:39]Option A is the right answer guys. Can I move forward?

[01:32:44]Guys, can I move forward? 90 ways is the right answer. Moving forward guys. A bag consists of guys. A bag.

[01:33:03]A bag contains four red, three black, two white balls. In how many ways? Three balls. In how many ways? Three balls can be drawn from a bag so that they include at least one black. Now Aman sir if we solve something like this I will tell you both the ways at least one is black.

[01:33:28]You are selecting three balls. At least one is black. What if guys one student can solve one is black and two are of different color. Two black one different color. All three black guys. In how many ways you can select one black out of three? 3 C1 multi two different how many different balls are left? 3 + 2 4 + 2 6 6 C2 guys. Two black 3 C2 one different 6 C1 all three black 3 C3. But Aman sir this will take lot of time. So guys I will give you one shortcut. Amen. What if we do like this total ways of selecting three balls minus the ways when no one is black? No ball is black. When no ball is black.

[01:34:18]Guys, in how many ways you can select three balls out of 4 + 3 + 2 out of nine balls guys? 4 5 6 9 out of nine answer we can select 9 C3 is the total ways of selecting three balls. Now at least one should be black. If you subtract the ways when there is no black, it will be the cases will left will be at least one black. No black means entire selection is done from this two. 4 + 2 6 minus 633 63. Let's do it guys. 9 into 8 into 7 divided by 6 84 - 6 into 5 into 4ide by 6 - 20 everybody.

[01:35:03]84 - 20. Are you all getting 64 is the right answer?

[01:35:09]Yes guys, 64 is the right answer. Yes, 64 is the right answer everybody. Option A 64 is the right answer. Is this point clear to all of you? Guys, is this point clear to all of you? Can I move forward sir? Can we do 8 C3?

[01:35:31]Why 8 C3? As one is compulsory, not one is compulsory. It's not like that. At least one should be black. At least one black is also the can be also the answer. Two black can be also the answer. Three black can also be the answer. At least one. One compulsory does not means that you have selected.

[01:35:48]No, no, no, no, no, no, no, no. You have to select. There are three different black balls. You have to select one. 3 C1. No, no, no, no, no, no, no. Who? No guys, everybody. Now moving forward. Now the maximum number of point of intersection of 10 circles. See guys they should use the word 10 different circle then it would be a better question 10 different circle. Ammon see whenever two circles intersect the point of intersection maximum point of intersections can be two. If the all the circles are of the same size the point of intersection will be infinite. So guys they should say the word of 10 different circles. But I am sorry guys but they have not mentioned amir in how many ways out of 10 you can select any two circles out of 10 you can select in this way you can select any two circles whenever two circles come we have two point of intersection these are the number of ways you can select any two different circle guys and having different circle whenever any one two circles come we have two point of intersection so number of circles multiply by these are the number of which you can select any two circle Whenever you to get two circles you get two point of intersection. So point of total intersection will be 10 into 9 divided by 10 into 9 / 2 45 * 2 90 ways.

[01:37:18]90 ways 90 ways. Now next question guys. There are five books of English, then four books on Tamil, three books on Hindi. In how many ways these books can be placed on a shelf? If the books of the same subjects are together, guys, box concept, put them in one box, put them in one box, put them in one box. Should I count this box as one unit? Yes, I'm answer three factorial. Even if they are arranged inside the box, still they are together.

[01:37:56]So guys, what is the answer? 6 into 120 into 24 into 6. Guys, are you all getting 1 0 3 6 8 0 1 0 3 6 8 0 guys? 1 0 3 6 8 0 Everybody guys, is this point clear to all of you? Yes. Ammon sir, a room has 10 doors. A room has 10 doors. In how many ways a person may enter from one room one door and come out of guys see indoor this is entry door and this is exit door entry and exit indoor and outdoor guys I'll write outdoor guys in how many ways a person can go into the room 10 C1 and means multiply he should come out of the different door for example he has taken door A now the door A is not available now out of nine in how many ways he can choose any one Guys, 10 into 9 everybody.

[01:38:55]90 ways. 90 ways everybody. Option A is the right answer everybody. Guys, option A is the right answer.

[01:39:10]Guys, can I move forward everybody?

[01:39:12]Guys, can I move forward?

[01:39:15]Guys, can I move forward?

[01:39:18]Yes. Now in an election there are guys in an election there are five candidates there are five candidates consisting of three vacancies guys there how many vacancies are there three an elector can vote the number of candidate not exceeding the vacancies there are three vacancies maximum you can give the vote in the maximum number of vacancies are there are three vacancies it means maximum you can give three votes okay in how many guys out of five a voter can vote either anyone 5C1 or he can give any two vote or he can give maximum three votes. Here the 2 to the^ n will not work because it is not extended till 5. 5 C1 will be 5 5 C2 5 into 4 / 2 10 5 C3 5 into 4 into 3 / 3 factorial 6 10 25 is the right answer now can I move forward everybody body.

[01:40:30]Now guys, a box contain ammon sir. A box contain a box contain everybody. A box contain how many things? Seven red, six white and four blue balls. How many ways selection of three balls can be made? So none is red. As you are selecting three and no one should be red. It means entire selection should be done from this 6 and four guys. What is 6 + 4? 10. Out of 10 ammons we need 10 C3. Three balls. If the selection is made three balls of the out of this there is no one is red. So what is the answer? 10 into 9 into 8 divided by three factorial that is six. Everybody 120 ways is the right answer. Yes. Yes. Yes. 120 is the right answer. Guys, I hope you are getting the confidence while solving this questions. The question paper comes very easy. You just have to apply presence of mind. Apply the NPRNCR and get the answer. Moving forward, in how many ways? Five different trophies can be arranged in a shelf. If one particular guys, one particular trophy must occupy the middle place. Let's see.

[01:41:48]1 2 3 4 5 five trophy T1 T2 T3 T4 T5 they are saying one particular I don't know particular means it is specified for example I'm talking for about T1 T3 it must occupy the middle place now one place is gone one trophy is gone now how many place are left 1 2 3 4 in how many ways five it's just like one particular person T3 come you will take this place now T3 is not part of my selection not part of my arrangement. Now in how many ways? Four trophies can be arranged in four places. 4 P 4 factorial that is 24.

[01:42:28]That is 24. Yes guys. Yes. Yes. Yes.

[01:42:31]Correct answer everybody guys. Yes.

[01:42:33]Correct answer.

[01:42:39]Moving forward.

[01:42:41]Now guys a very good question. Mark this question as an LDR. Everybody mark this question as an LDR.

[01:42:52]Mark this question as an LDR. Now there are six men's and four women.

[01:43:02]There are six men and four women in a group. The number of ways they can form a committee of five. They can form a they can form a committee of five so that it includes at least two women. Come at least two women. Amir out of at least two. If someone will say can we do like this total waste minus ways when there is zero women when there is one woman. Yes you can do but in that case also you have to solve three cases.

[01:43:33]Total ways minus ways when there is zero women when there is one woman. If they would have said at least one woman total ways minus ways when there are no women but they're saying at least two it is better let's try to draw the cases since we have to make a selection of five ammon can we say like this case one two women woman three men can we say like this ammon three women one two three and guys two men Aman sir case one more case sir four women 1 2 3 4 and one man one more case sir all the five are women but the problem is we are having only four women's available so we can't draw the five women case let's solve it guys let's solve it we need two men out of four sorry we need two women yes women two women four C2 we need three men out of six C3 3 Three women 4 C3 two men out of six 6 C2 All four women 4 C4 1 men out of 6 C1 let's do it guys 4 C2 let's do it 4 into 3 / 2 6 6 into 5 into 4 / 6 this into MRC guys are you all getting this is 120 next one 4 C3 guys 4 into 3 into 2 / 6 m + 6 into 5 by 2 this into MRC are you all getting the 60? Yes. 4 C4 is 1 6 C1 is 6. How much is the total? 120 + 60 + 120 + 60 + 6. Are you are getting guys 186?

[01:45:25]Yes. I'm answer 186. Everybody option B is the right answer. 186 option B is the right answer. Is this point clear to all of you guys? Is this point clear to all of you? Can I move forward guys? Is this point clear? Very good question. Now in how many ways can a selection of six? I have to make a selection of six. Out of how many? Six teachers and eight students. Six teachers and eight students. Again the same based question. You have to make a selection of six so that it includes at least two teachers guys at least two teachers. Two teachers three teachers sorry there are not that's why I was worried four teachers sir out of six at least two at least three four maximum three cases. So I will draw the three cases. Case one at least two teacher sir teacher teacher you have to make a selection of six two are at least teachers how many will be students four students case number two among s at least two means there can be three teachers also rest three will be students one more case all four teachers and two students among sir if they would have given five teachers five teachers one student like that but since the teachers are only four maximum This can be drawn now.

[01:46:56]Yes. Yes. Yes. Guys, now let's do it.

[01:46:59]Out of four, you need two. 4 C2.

[01:47:03]Out of eight, you need four. 8 C4.

[01:47:06]Guys, you need three teachers. Out of 4 C3, you need three students. 8 C3. Four teachers, guys. 4 C4. You need two students out of eight. 8 C2. Let's do it, guys. Let's do it. Yes, let's do it.

[01:47:24]4 C2 4 into 3 / 2 M + A 1 2 3 4 / 24 this into M RC guys this you will get 420 A4 C3 guys 4 into 3 into 2ID 6 M + 1 2 3ide 6 this into MRC guys you must be getting 224 4 C4 is 1 8 C2 8 into 7 / 2 you must be getting 28 8 guys let's get the answer everybody 420 + 224 + 28 are you all getting 6 guys are you all getting 6 72 everybody yes this is the right answer 672 is the right answer guys are you all getting the answers everybody is this point clear to all of you yes is this point clear to all of you yes now Guys, there are if there are 40 guest in a partying, how many guests? 40 guests.

[01:48:28]The if each guest shakes hands with one another remaining guest, guys. To have one handshake, we need two person.

[01:48:36]Having two persons means having one handshake. So out of 40 in how many ways you can select two persons? Because having two person means having one uh having two person means having one handshake. The number of ways you can have two persons is equals to number of handshake. So guys 40 into 39 divided by two guys there will total 780 handshakes. Sometimes they give you number of handshakes is this much. You tell me the number of person. So we'll go from option to question. The option that will give me this much number of handshakes that option will be correct.

[01:49:10]Now if why not we can do 40 P2 why this is wrong?

[01:49:18]Now see either I shake hand with Ram or Ram shakes hand with me. Both are same.

[01:49:25]It should not be counted as two handshakes. Since order is not material here, I will not uh say arrangement. I will not say P. Now next question. In a basket guys, in a basket there are seven apples, six banana and four mangoes. How many ways the uh selection of three fruits can be made? All three are apple. All three are apple. It means entire selection has to be done from the seven.

[01:49:56]Rest they have just given the information to manipulate you. 7 C3 you have to make a selection of food only. 7 C3. 7 into 6 into 5 / 6 you must be getting 35 ways guys. You must be getting 35 ways. Yes.

[01:50:15]Moving forward guys, now again the question based on the same scenario guys. You are having seven boys and four girls.

[01:50:27]Seven boys and four girls. A team of guys, a team of five. You have to make a team of five is to be chosen. The number of teams such that each team includes at least one girl. Since it is in at least one one girl, two girl, three girl, four girl, four cases, can we have one shortcut? Yes. I'm answer do like this.

[01:50:49]Total ways of selecting five total ways minus when no one is gull. No one is gull. If out of total ways if you subtract the when when there is no girl.

[01:51:00]So number of ways left will be having at least one girl. In how many ways you can select any five out of 7 + 4? Sir 11 C4 minus when no one is skull. When no one is g it means entire selection is done from this seven 7 C4.

[01:51:22]Let's get the answer. 1 2 3 4ide by 24.

[01:51:27]As are you getting this 330?

[01:51:31]1 2 3 4ide by 24. As are you getting minus 35?

[01:51:38]As are you all getting 29? Something is wrong.

[01:51:42]So sorry guys. Selection of five.

[01:51:44]Selection of five. So sorry guys.

[01:51:45]Selection of five you have to make. So sorry guys. You have to make a selection of five. 11 C5. 7 + 4 guys. Yes. 11 C5 minus when no one is girl all are boys - 7 C5 so sorry my mistake 1 1 2 3 4 5 divided by 120 yes 462 1 2 3 4 5 divided by 120 as you must be getting 21 462 minus 21 you must be getting as 441 you must be getting 441.

[01:52:24]Guys, is this point clear to all of you?

[01:52:26]441 is the right uh answer. Guys, can I move forward? Everybody guys, can I move forward? We are almost at the verge of completing this chapter. Now guys, there are n locks and n uh corresponding keys.

[01:52:41]The maximum guys here the word is the maximum the maximum number of trials are needed to assign the key to the corresponding locks. corresponding logs guys I'll give you within a shortcut scene there are four locks let us assume there are five logs let's say guys let's assume there are guys let's assume there are five locks and how many keys guys there are five keys now maximum number of trials Then I will give a try to this one. How many trials I will make? Let's think and say how many maximum trials. Maximum trials is possible when only last key will satisfy this lock. Only when the last key will satisfy. So you will try four keys. I will not even try fifth key. I say since four key fails. Now fifth key is the key of this lock. So guys, I will say maximum trial will be four here. Now one key is gone and one lock is gone.

[01:53:47]Now guys you are having four keys left to try this one you will not you are having four keys now you not try the four key maximum you will try three key that will not satisfy so you say this last key belongs to this lock you will try three three keys in the same way guys you will try one key and guys here you will try one key guys for last one you will not have a zero try why because one key is left you say this key belongs to this even I will not try so what is the maximum number of tries 4 + 3 + 2 + one answer we will have maximum 10 trials 10 ways.

[01:54:20]Now let's go from option two question guys. If you consider answer 4 5 - 1 4 4 C2 4 into 3 divided by 2 you are getting 6 wrong. Let's put n over here 5 + 1 6 c2 6 into 5 divid by 2 you are getting 15. No we are searching for 10. Let's put it over here. From 2 to n what is your n? 5. So guys when we will put two over here 2 - 1 1 + you will put three 3 - 1 2 When you will put three guys 3 - 1 guys 4 4 - 1 3 When you put guys till n is your five 5 - 1 + 4 1 + 2 + 3 + 4 I'm answer this options give me 10. So everybody option C is the right answer.

[01:55:11]The best way is to assume let the number of locks be five and number of keys are five and you have to apply the common no need of any permutation combination. No, this was a common sense based question more maximum trial. If they would have said minimum trial guys my first key will satisfy this first key will satisfy this first key will satisfy this. I will satisfy one way, one way, one way, one way. Total 1 + 1 + 1 plus 1. Minimum T.

[01:55:36]My first key will satisfy. Maximum T last key will satisfy. You are having five keys. So we'll try four keys. Then say fifth key belongs to this. So maximum trial for to get the correct key for this is four trials, not fifth one.

[01:55:47]I will not check fifth one. I know this is the answer. This will satisfy. Now everybody guys, moving forward in how many ways? Four doctors, five doctors, four professors and six auditors be seated in a room so that the persons of the same professions are together. I will put five doctors in one box. Madam, four professors in one box and six auditors in one box. Since you want them together, count them as one unit guys. I will say three factorial. Even if they arrange inside the box, still they are together. Multiply by 5 factorial * 4 factorial mult* by 6 factorial.

[01:56:29]Everybody guys you can check. Option C is the correct answer. Yes, option C is the correct answer.

[01:56:37]Yes, everybody guys with me. Everybody guys with me. We are done with the question number 61. Now in how many ways can an interview panel of three members be formed out of three engineers, two psychologist and three manager. You have three managers. What is the question requirement in how many ways? A interview panel of three. You have to select any three. You have to select any three. Okay. It includes at least one.

[01:57:12]At least one out of three one can be engineer, two can be engineer, three all the three can be engineer. You can draw three case one engineer, two other, two engineer, one other. All three engineer answer. But we have the better way.

[01:57:27]Total ways minus when no one is engineer. When no one is engineer in how many ways? Total ways you can select three. How many persons are there? 3 + 2 + 3 amounts are 8 C3 out of this is the total ways of selecting three persons when no one is engineer it means entire selection is done from this 3 + 2 5 - 5 C3 guys 8 C3 do it guys 8 into 7 into 6 divided by 6 M + 5 into 4 into 3 divided by 6 is equ= to M - MRC guys you must be getting 46 6 you must be getting 46 everybody. Option C is the right answer. Option C is the correct answer guys. Yes. Yes. Yes.

[01:58:18]Everybody. Is this point clear to all of you? Guys, can I move forward now? Guys, the algebraic based question I have drafted a very easy one. You have to solve it by yourself. Everybody know this thing guys. Every Wait, wait, wait just a second.

[01:58:55]Just a second guys.

[01:59:05]Guys in this let me make you revise two things. If two combinations are same, first r must be equals to s and second this plus this must be equals to n. So if you use this guys, since two combinations are same, two combinations are same. This plus this must be equals to this. So you will say 3 r + r + 3. This must be equals to 15. 4 r must be equals to 12. 3 will go there guys. So you will say r must be equals to 4. 3 are 12. Everybody guys R value is three. Yes, everybody R value three.

[01:59:41]Option B is the correct answer. In the same way guys this you have to go from option to question the option which will give you 45 that option will be correct guys. The option that will give you 45 will be correct. Let's check for option B. I'm answer if you check for option B.

[01:59:57]Let's check 8. 8 + 2 10 C8. Let's solve 10 C8 guys. What will be 10 C8 guys? n factorial upon n - r factorial into r factorial. So guys 10 into 9 divided by 2 am this option give me 45.

[02:00:19]This options give me LHS equals to rhs.

[02:00:23]So this option is correct. Rest all the options are wrong. So guys in this case you have to go from option two question.

[02:00:30]Moving forward guys. NPR upon NCR. I have given you what this guys what is the ratio of NPR upon NCR. This is the R factorial. Amir if we do like this everybody knows NPR upon NCR. This is R factorial. In short 336 guys in short 336 divided by guys 56 this is equals to R factorial. So guys what will be r factorial 336 divided by 56 amans sir we are getting six we can't say r is 6 they are saying and r don't say r is 6 am sir from here we can say r guys from here we can say r must be equals to 3 r must be equals to 3 so that the factorial will give you six so guys the option which is having r as three here r is 3. This will be correct.

[02:01:31]Here R is 22. Here R is four. So guys, even you can put the value of R over here and you can check whether 8 C3 gives you 56 or not. Guys, everybody, option B is the correct answer. Even if you forget this guys, you can just go from option to question. The option that will give you 56 and the option which will give you 336, that option will be correct.

[02:01:54]Huh? 3, 6.

[02:01:58]No. Turkish wrong not 3a 6 guys just a second 8a 3 8 into 7 into 6 divided by 3 uh 6 guys 8a 3 r factorial is 6 and I'm not asking r factorial I'm asking r if you want factorial outcome six the r must be three because 3 factorial will give you six guys everybody with me yes moving forward guys you have to find the value of n what is the lcm guys here The LCM of the largest number factorial of the largest number is LCM. So how you will solve this everybody?

[02:02:36]So 8 factorial will be your LCM. N upon 9 factorial. Let's do it. 8 factorial divided by 7 factorial guys. 8 factorialide by 7 factor thing guys. 8 into 7 factorial divide by 7 factorial cancel 8 * 1 8 + 8 factorial divide by 8 factorial 1 1 into 1 1. So guys you will get 9 upon 8 factorial n upon can I say 9 into 8 factorial this will cancel. So guys from here you will get guys from here you will get uh guys from here you will get n= to 9 into 9 that is 81 everybody guys 81 81 yes guys is this point clear to all of you guys one question on the pascal law I have told you whenever they ask this question on pascal law I have taught you guys very well see ncr plus n c r minus one must give you n + 1 c. So the whenever they ask the question on this go from option to question the option which will satisfy the pascal law of that optional is correct. Now it is very easy to decode here see n we need here n must be same then only we can say n + 1. So everybody know that here n must be 999. But the problem is I'm answer even when we put option A we get 999 here we will put okay I'll write like this guys I'll put over here RS 999 C97 plus in the place of X if you put option A 901 am we can say this N + 1 C R only when this is one more or one less we are not having difference I told you guys in the class also that this and this works in conjunction.

[02:04:32]If option don't match, you can replace n c from n c n minus r. If you apply that formula, if you split this n c n - r 999 - 901 you will get this as 98. I'm just ncr and n c n minus r both are same. N C R N C N -R R plus I'm answer this is same 999 C97 so what you must get n + 1 n + 1 is 1,000 1,000 cr r bigger number what is bigger number this is r this is r now this is r minus one 98 yes sir this option is give you ls equals to rhs so this option is correct rest all the options are wrong and guys here we are done with the discussions of all the variety miscellaneous word number geometry circular everything guys grouping distribution formation group everything guys here we are done with all the variety of the problems and here we conclude the practice session of the chapter extra practice all the variations as you know chart book and this it is just as a regular batch no need to solve any pyq and genie book anything guys if you're done with this you can stop this much is sufficient See you guys in the next class. Bye everyone.