Day 25: Stars and bars agaaaain! This time with the restriction that one of xi’s is at most k.

Ajouté : 2026-06-07

[00:00:01]Let's say you go to a bakery shop and you want to buy exactly 10 croissants and there are four different types of croissants: plain, butter, chocolate, and almond croissants. What you know is that you want to buy at most three chocolate croissants, which means that you're not allowed to buy four or five or any number bigger than that. But there is no other restriction on the other ones. They're all greater or equal to zero. You can map this problem to this equation: X1 + X2 + X3 + X4 is equal to 10 while X3 is smaller or equal to three. We want to know how many solutions exist. One way of solving that is that since it is at most three, you consider the case that you buy zero, you buy one, you buy two, and you buy three.

[00:00:47]Which means that if we buy zero, then the equation reduces to X1 + X2 + X4 is equal to 10. If we want to buy exactly one, the same equal to nine. If we want to buy exactly to two, the same exactly equal to eight. And if we want to buy three, it is going to be X1 + X2 + X4 equal to seven. And in the previous videos, we learned how to solve each of them and then we can add the numbers here. But is this solution going to work if you want to buy 1,000 croissants? And let's say that you have this restriction that the number of chocolate croissants is less than or equal to 100. Is it easy to go with exactly zero, exactly one, all the way to exactly 100? I don't think that it is a good solution. In the next video, I'm going to talk about a better way. If you want to learn that, please subscribe. This is day 25 of my 30-day challenge in counting and combinatorics.

[00:01:46]See you in the next video.