53 Probability Part 1 Basics of Permutations and combinations 1

Hinzugefügt: 2026-05-06

[00:00:00]hello and welcome to SN Gate Academy our next chapter is probability now before starting probability we will learn some of the basic concepts of permutation and combination now we will start the permutation combination with fundamental principle of counting now in fundamental principle of counting basically we have two different rule the first rule is multiplication rule and the second rule is addition rule now this multiplication rule says that if a work a can be done in M different ways and work B can be done in n different ways and C is a work which is completed when both a and B are completed okay so this is the statement of multiplication rule there is a verdict this is completed in to a step the first step is a and the second step is B the first step can be done in M different ways and the second step can be done in n different ways now the multiplication rule says that if the work C is completed when both a and B are completed then number of ways of doing this work C this will be simply M into n so let me give you on a very basic example suppose if there are three points a B and C and our job is we have to go from point A to C but the first step is I have to go from A to B and the second step is I have to go from B to C now suppose there are two paths connecting point A and B there are two ways to go from point A to B either you can go along path one or path 2 and say there are 3 ways to go from point B to C say these are the three paths connecting point B and C these are the three paths ABC now here if I ask you how many ways are there to go from point A to C so the first part of the job can be done in two different ways the second part of the job can be done in three different ways so the number of ways of doing this work this will be simply 2 into 3 6 so this is a very basic jumble of multiplication rule that if a work is done in number of steps then number of ways of doing the total work that will be the product of individual number of steps you can take another example suppose if I have four points say I have the point a B C and D and I have to go from point A to D but the first step is I have to go from A to B now say there are two ways to go from A to B then second step is I have to go from B to C and there are three ways to go from point B to C and similarly say there are four ways to go from point C to D now here in this case if I ask you how many ways are there to go from point A to D so A to B this is the first step this can be done in two different ways B to C second step this can be done in three different ways and C to D this is third step and this can be done in four different ways so here the number of ways of doing this total work which is going from point A to B this can be done in two in two or three into four that is to it that is 24 is so you have to apply this multiplication rule when the work is completed in number of your steps and the total work is completed when each of the individual steps is completed so this was first rule multiplication rule the second one is it is same rule okay the next one is addition rule now the statement of this additional is similar to multiplication rule the statement is that if a work a can be done in M different ways and what B can be done in M different ways can see is a work which is completed when either A or B is completed now notice the difference here in case of multiplication and all the work C was completed when both a and B is completed now here in a decent rule the work C is completed when either a or b is completed then in this case the number of ways of doing this work C this will be simply M plus M so again let me give you an example suppose if I have a rule and say there are two gates on left side these are the two gates one and two and say there are three years three gates on right side say these are the biggest one two three now here if I ask you how many ways are there to enter inside the room so you can enter either from the left side or from the right side there are two ways to enter from left side three ways to enter from the right side so here the new number of ways of doing the but this will be simply to +35 now here we have applied the additional because the work is completed in only one a step either you can enter from left side or from right side so when a work is completed in number of steps then you have to apply the multiplication rule and if the work is completed in only one step then you have to apply this a decent rule now we are going to see some of the basic problem and with the help of this we'll see how when to apply this radiation tool and when to apply the multiplication rule so our next column is number of four digit even number the number of four digit numbers that can be formed using these five bridges now notice this problem is similar to the one that we have already solved except that I have one mold is it a zero here okay these two are same first part when repetition is not allowed second part when the days can be repeated any number of times now again four digit number means we have to fill these four places using the given five digits but now we have a problem the problem is that we have this zero here and we cannot place zero on thousand place okay if you play a zero in thousand place then in that case the number ability of three days it the question is we have to form four digit number so this means we cannot place zero on thousand place so we have to start with the thousand place in the earlier example we have started with unit place but we could have started from any other place also and good off we would have obtained the same answer but here since there is restriction on thousand place because we cannot place zero on thousand place so we are going to start with this thousand place so how many ways are there to fill this thousand place so you can see that there are four nonzero degrees one two three four and you can place any one of these four digits on thousand place so this means there are four ways to fill the thousand place either you can place one or two or three or four now since the repetition is not allowed for the next place total five digits are there one of the digit is a thousand place so for the next place we will have four options okay five minus one one is one thousand place so for the next place will have four option because one non zero is here but on hundred place we can play zero here so there are four ways to fill this one then for 10th place total 5 bridges are there two are at one hundred and thousand place so for the next place we will have three opsahl and for unit place will have no different option so in this case the number of four-digit numbers that will be 4 into 4 into 3 + 2 2 so this will be 24 into 496 so men's ero is there and repetition is not allowed then total 96 numbers are possible using these fibrils and the second part of the problem is when the disease can be repeated any number of thing so now again we have to fill these four places using the given five periods and since we have zero here so we have to place non-zero digit at thousand degrees so there are four ways to fill this thousand requests we can place any one of these four nonzero digits on thousand rupees then 400 place will have five different option since the repetition is allowed any digits can be repeated any number of times so there are five option for 100 days similarly five of sand for 10th place and five of cell for unit place so when the repetition is allowed then in that case the number of five digit number will be 4 into 5 and 2 5 into 5 so this will be 125 into 4 that is 500 so when the petition is not allowed then 96 numbers are possible when repetition is allowed then 500 numbers are possible okay so our next problem is find the number of five digit event numbers that can be formed using the dearest 1 2 3 4 5 when repetition is not allowed so here we have to form even number now we know that a number is even number when there is it on unit places even so first of all we have to form 5 digit number so this means we have to form 5 box so say these are the 5 box now since we have to form even number so therefore we have to start from the unit place see in earlier example we started with thousand lists because the reason was we cannot place 0 on thousand place we had to place any non-zero digit one thousand place now here the problem is we have to form even numbers so this means we have to start from unit place now the restriction is on unit place we have to place event is it on you so now you can easily see that that there are two event is is two and four so this means there will be two ways to fill this unit pace either you can place two or you can place for on this unit place and the remaining box you can fill in any order you can so since the repetition is not allowed and one of the digit is at unit place so number of ways to fill the tenth place this will be four then three two and two one so number of five digit even numbers that can be formed using these five digits this will be simply 1 into 2 into 3 into 4 into 2 so this will be 48 now this was a very simple problem next I'm going to solve ok so this was very basic problem now next we are going to see another variation of the similar problem so our next question will be find the number of five digit even numbers that can be formed using the digits and okay so our next problem is and this is a very interesting problem the question is find the number of four digit even numbers that can be formed using the day.this 0 1 2 3 4 5 now see here we have to form an even number and 0 is also there so first let me explain what is the approach of their student and then we'll see what should be the correct approach so the question is find the number of folders it even numbers so this means we have to form we have to fill these four places now event number means we have to form we have to place event is it on unit place now the event is is are 0 2 & 4 these are the 3 event digits so suppose if I am writing 3 here 3 will mean that I'm placing 0 here or 2 or 4 so now since the number has to be a whole digit so this means on thousand days we have to place any nonzero digits so if I ask you how many ways are there to fill this thousand place so that will depend on what digit we have placed on unit place right if I have placed a zero on unit place so this means in that case four thousand place will have five different ops I can place any one of these five bridges on thousand place but if you have placed two or four on unit place so in that case you will have only four opps and because zero you cannot place on thousand place one of the non-zero digit either two all four is that unit place suppose if you are placing two on this unit place so this means four thousand place you will you cannot play a 0 2 is that unit place so this means four thousand place you will have only these four option either one three four or five so the number of ways to fill this thousand place that will depend on what digit you are placing on unit place if you are placing a 0 then you will have five option if you are placing two or four then you will have three then you will have four different option so for these type of problem we have to consider two different cases here you cannot write three because the moment you write three we don't know what we should write on thousand place the number of ways to fill this thousand place this will change depending on whether you have placed zero or nonzero here so for these type of problem we have to consider two different cases okay so the first case will be case 1 when 0 is at unit place 0 is at unit place so when 0 is at unit list then in that case since we have to form 4 digit number this means we have to fill these four places and it says 0 is that unit base so this means there is only one way to fill this unit this we have to place a 0 here now number of ways to fill this thousand place this will be 5 because you can place any one of the remaining 5 non-zero digit on thousand place and similarly number of ways to fill this 100 place total sex disease are there one at unit one at thousand so number of ways to fill this 100 place this will be 4 because you can place any one of the remaining four digit 100ms and the number of ways to fill the 10th place this will be 3 so when 0 is that unit bless this make it when 0 is at unit this then in that is the number of even numbers will be 5 into 4 into 3 & 2 1 now if you multiply all these numbers 20 into 360 so 60 even numbers are possible which will have zero at unit place then second-guesses case 2 when 2 or 4 is at unit place now when 2 or 4 is at unit place again we have to fill these four places when 2 or 4 is at unit less then in that case there are two ways to fill the unit place either you can place two here or you can place for here so there are two ways to fill the unit place and now we have to fill this thousand pairs so the number of ways to fill this thousand place will be you cannot play a zero here one non-zero digit is already at unit place so this means number of ways to fill this thousand place this will be four okay so you know you can't place one of the nonzero digits is at unit place so out of these five nonzero digits only four are there for this thousand place so number of ways to fill this thousand place this will be four then for the hundred place again we had total digital six two digits are at unit and thousand place so number of ways to fill this 100 place this will be four and four tenth place you will have three options so in this case the number of even numbers will be 4 into 4 into 3 into 2 now if you multiply all this then you will get 96 so 16 of us 60 even numbers are possible which will have 0 at unit please 96 even numbers are possible which will have 2 or for that unit place so the number of even numbers will be 60 plus 96 so this will be 156 notice that here we had applied the addition rule because here the work is complete either you can form even numbering in this manner or in this second case so number of even numbers will be 60 plus 96 misses one of your basics so this was a very good problem ok now our names Walla means find the number of four digit numbers divisible by four that can be formed using the digits one two three four five when reputation is not allowed so we have to form four digit number which are divisible five now the condition divisible by 4 so the condition for divisibility by food is that the last two digits would be divisible by 4 okay for example suppose if I have a number say 4 6 7 1 2 now if I want to find whether this number is divisible by 4 or not then you have to see only in the last two digit if the last two digit is divisible by 4 then in that case this number will be divisible by 4 so here I worked all the male's we have to form now obviously this number is will be divisible by 4 because 12 is divisible by 4 now our problem is we have to form four digit numbers divisible by four that can be formed using these five digits so since we have to form four digit number we have to fill these four places but now we have to fill the last two places simultaneously the reason is I have to place only those pair of number four in the last two places which are divisible by four so we are going to see what are the possible options for the last two place so if I have one on tenth place then I can place two on unit-based so one possibility is one two I can test one two on these last two blesses I mean that case the number will be divisible by four so one possibility is going to one to another possibility will be to 4c you don't have to write these combinations randomly you have to proceed in a sequential manner if I have one or tenth place then I can place two on unit place if I have to one tenth place then I can place for on unit place similarly three two okay then if I have four on tenth then I should have four on unit place but since the repetition is still not allow that I cannot place for four so nothing with four then next if I have five on tenth place then I can place to one unit place so these are the four possibility for the last two places either you can place one two here all two four or three two or five two so the number of ways to fill the last two places this will be four okay there are four ways to fill the unit place and 10th place simultaneously sleep in which the number will be divisible by four now the number of ways to fill this tenth 100 place total five places are there two digit is at unit and tenth place so this means remaining number of days it is three so number of ways to fill the hundred place this will be three and the number of ways to fill the thousand place this will be two so here the number of whole digit numbers divisible by four this will be simply 2 into 3 into 4 which is 24 okay so this was a very basic problem only thing is that you have to see that you have to fill the last two places simultaneously only in that case the number will be divisible by 4 so these were some of the basic problem based on the fundamental principle of counting now next we are going to see permutation and then we will talk about combination okay so our next topic is permutation now the meaning of this word permutation is arrangement permutation simply means arrangement and the first theorem is the number of permutation of n different object now this word different is very very important we can apply this theorem only when all the objects are distinct or different so the number of permutation of n different things taken are at a time is given by now the formula is NP R this will be the number of ways to arrange are out of an object so this NP R will be simply factorial n this divided by factorial n minus R now with the help of one example we will see how to apply this formula so for example suppose if the question is find a number of four lettered word find a number of four lettered word that can be formed using the letters of word that can be formed using the letters of world machine so this is the problem that we have the Versa yes we have to form for later world and that we have to use the letters of this word machine now the question is what formation is an example of permutation okay here we have to arrange for out of these seven of letters so total seven letters are there so this means n 8 7 M SCH i ne so total 7 letters are there so this means n 8 7 now the question is we have to form four lettered word so this means in place of our you have 4 10 8 7 RS 4 so the number of four lettered word this will be simply 7 P for now 7 P for you can write as factorial 7 by factorial 7 minus 4 so this will be factorial 7 by factorial 3 now factorial 7 you can write as 7 into 6 into 5 into 4 into 3 factorial and this divided by 3 factorial now if you do the calculation this will be 42 into 20 so this will be 840 so this means total 840 words are possible which are all related and which can before and the and the letters that are used or from the letters of word machine now see here we can apply this formula NPR because all these seven letters are distinct okay if there is some repetition here then you cannot apply this formula so therefore this theorem is to be applied only when all the N objects are distinct now this was first theorem the second one is the number of permutation of n different objects taken in at a time okay this is second theorem so number of rotation of n distinct object taken in at a time this is given by NPN and this NPN will be nothing but factorial n by factorial n minus M now factorial n minus n will be factorial 0 factorial 0 is 1 so this will be factorial n so number of ways to arrange any object on in places this will be simply factory and see the value of factorial 0 is defined as 1 1 lead to make this formula NPR a valid when R is equal to n we have seen that the formula for NPR is factorial n by factorial n minus L now when R is equal to n then in that case we know that n PN will be simply factorial n so one way to make this formula valid when R is equal to n factorial 0 is defined as 1 so these are the two basic Europe that you have to remember NPR will be factorial n by factorial n minus R NP n will be factorial n this formula you have to apply only when all the N objects are different now we are going to see some of the special cases in permutation so our next heading will be permutation under different conditions okay so our next next topic is permutation under different condition now under this topic we'll see different type of cases so the first case is when all objects are not distinct see the formula for NPR that we had seen was factorial n by factorial n minus 1 and the formula for n P and this was factorial n but these two formula you had to apply only when all the N objects are different now here we will see what happens when some of the objects are same so I will explain that concept with the help of one example suppose if I have this word a BB L and here if I ask you how many five letter word are possible using these five letters so here you cannot apply this formula NPN the reason is that all the N objects are not different you have two identical P now for a moment consider that these two letters are different one letter is P 1 and another letter is P 2 now in this case if I ask you how many five letters are possible so now we have five letters and all the five letters are different when is P one other is P two so in that case the number of vibrated what this will be factorial 5 right factorial 5 this will mean simply 120 so now what are those 120 word first what is AP 1 P 2 le the second word will be a P 2 P 1 L the third word you can think of is al e p1 p2 another possible word is al e p2 p1 and in this way you can form total 120 word now see p1 p2 are not different ok so this means these two words are same the first word is a double ple the second word is also same a double ple so these two words are same similarly these two words again her same al e p1 p2 and al e p2 p1 so total 120 words are there out of it these two are same these two are again same so this means the number of different word this will be simply factorial 5 by 2 what you can write this as factorial 2 ok so when some letters are you did then you have to divide by the factorial of that number so here the number of different words this will be simply factorial 5 by factorial 2 now you can think of another example ok to understand the concept of this meaning factorial suppose if I have a word say D a double D Y now here if I ask you how many five letter word are possible and say if one is t1 other is d2 and another one is d3 so in this is the number of different words that is possible is again factorial 5 we have 5 later and all the five letters are different now factorial 5 this will mean 120 so total 120 words are possible now what are those words one word is d1 d2 d3 why another word you can think of is d1 a d3 d2 y similarly D 2 a d1 d3 y d2 a d3 d1 y d3 a d1 d2 Y and finally d3 a d2 d1 y so in this may total 120 words you can form now if you see carefully these six words are same all the words are d a double D Y so total 120 words were possible out of which these factorial three number words are same another factorial three number of words are same so this means the number of different what this will be simply factorial 5 by factorial 3 okay so this was very basic example when all the objects are not distinct then you have to divide by the factorial of number of times any particular letter is repeated another example that you can think of this and this is a very popular example and then this after is incomplete without this example so suppose if I have this word say Mississippi and if I ask you how many 11 letters word are possible using these 11 letters okay you can think of you can easily see that there are total 11 letters so if I ask you number of a label later word that can be formed using the letters of word Mississippi so this will be factorial 11 this divided by now you can see that there are four identical s okay you have two s here and 2 s here so you have four identical s then four identical I to P and one M these are their level letters that you have so number of a level lettered word this will be simply factorial 11 this divided by factorial 4 then 4 identical eyes for factorial for two identical P so again factorial 2 so this will be the number of 11 lettered word that can be formed using the letters of what a Mississippi so whenever one hour modulator is repeated then in that case you have to apply this approach now so this was first type next we are going to see some other cases okay so next type is when certain types of objects are together so I will explain this with the help of this example the question is fine the number of ways to arrange the letters of word daughter such that all the vowels are together now see there are total eight letters you have three baubles au e and then you have five consonant DZ h T or so these are the five consonant and these are the three vowels now the question is we have to arrange these eight letters in such a way that all the three worlds are together so for these type of problem you have to count these three vowels as one single unit okay so you have to count these three vowels as and then you have five unit of these five consonant now number of ways to arrange these six letters okay five consonant plus 1 unit of these three vowels this will be factorial six okay and in all these arrangements you can also arrange these three wobbles in factorial three ways so this means the number of ways to arrange this will be simply factorial six in two factorial three now let me explain this thing further suppose if if one word is said Dez it's TR and then you have a ue another possibility that you can think of it's a d g th r then again a ue so in this way by taking these three by counting these three has one total number of ways to arrange these six units maybe you can think of and say au e bz HT r so in this way total factorial six number of words you can form these three as one and then you have these five units so total number of ways to arrange this six unit this will be factorial 6 and from each one of these words you can form six another word right by array by interchanging these words you have two d z h TR and then you can entertain say you and ii so you can write a ue so this will be one word so in this way from one word you can form six another word by interchanging these three letters so from each one of these factorial six number of words you can form factorial three number of words from each one of these words so the total number of different word will be factorial six in two factorial three now suppose if the problem was say this was the first part of the problem the second part of the problem is all consonant together we have to arrange these eight letters in such a manner that all consonants are together so now you have three vowels au e and then you have five consonants Dez STR and now after all the ways we have to arrange in such a manner that all the five horsemen together so this means we have to count these five consonant as one unit so you have one unit of these five consonant and then three unit of these three vowels so the number of ways to arrange these four units will be factorial four and in each one of these Arrangements you can also arrange these five consonants in factorial five ways so this will be the number of ways to arrange these eight letters such that all the consonants are together so this was again a very basic example now our next type is the third world