 Hi, I'm Zor. Welcome to Unizor Education. I would like to add one more interesting problem to combinatorics subject of this course. The whole course is presented on Unizor.com. I suggest you to watch the lecture from this website, because every lecture has a very good detailed explanation, notes. Plus, there are exams and some other stuff. So, the site is free, so I do recommend you to watch it from this website. Now, speaking about combinatorics, the most important part in combinatorics, I believe, is solving problems. Now, this is actually just yet another problem, but it has a very fancy name. It's called Wandermon's Identity. Now, sometimes it's presented like this. Okay, here is the identity. Basically, I wrote it for myself, so it doesn't really make sense. And now this is the proof. Now, just looking at identity, you don't really feel the combinatoric sense behind it. So, I would rather start from the meaning of that as an example of a practical problem. And basically, the solution to this would be actually the proof of the identity itself. And then I will offer another proof. Okay, so here is the problem. Let's consider you have a set of people of two different kinds. Let's say men and women or, I don't know, whatever, higher education and lower education. It doesn't really matter. So, you have M of, let's say, men and M of women. Now, your task is to create a subgroup which contains our participants. And the question is how many times, how many combinations you can create of our members of this group. Now, the obvious and very simple response is, well, we have M plus N people. It doesn't matter what kind of people we have. We have R, the capacity of the group which we have to select from it. And there is a direct answer, number of combinations from M plus N by R. And this is the answer. That's it. End of story. Now, on the other hand, you can approach this problem slightly in a more complex way. Namely, you can actually take into account that you have two different types of people. So, this group of R can contain zero men and R women. And there is a special number of groups of this type. This is what? This is number of combinations from N women by R. So, this is the number of these guys. Let me actually write it right near it. So, it will be C N by R. Now, another type of groups. When you have one man and R minus one women. The number of these groups is how many different selections of one man we can actually select from M men. It's C M by one, right? And how many different selections of R minus one women we can select from N women. It's this, right? Number of combinations from N by R minus one. Now, we can have two men and R minus two women, right? So, the number of combinations is this is number of times we can select two men out of M. And this is the number of times we can select R minus two women from N and we have to multiply because with each selection of this we can have each selection of them. Now, we can continue this process until R minus one and one would be C M R minus one times C N of one. And finally, R zero C M R C N zero. But I didn't put C M zero here. It's one, but just to be a uniform, right? Number of combinations from M by zero is still one, right? So, we don't have to take it into account. And now, my point is that you can summarize this and that would, well, it's supposed to give you exactly the same number as this one, right? We just count it differently. This is the direct count, regardless of the composition, and this is a count which does pay attention to the composition of the group between men and women, but we do actually select all the different kinds of composition. Zero men and R women, one man and R minus women, et cetera, up to R men and zero women. So, we have this type of numbers of groups of this particular type. So, if we will summarize them, we will get this. And this is the Van der Monde's identity. So, basically, I have just proven this identity by purely logically approaching this problem. And my personal understanding is that most likely this particular identity, which I can now express as this, C M plus N by R is equal to sum, and I will have index I from zero to R, C M by I, C N by R minus I. So, this is the sum of these expressed as a sigma notation. Well, that's it. That's the proof. This is the identity which we're talking about, the Van der Monde's identity. Again, very fancy name, but very simple explanation where it came from. And now, kind of elegant, I would say, maybe slightly unusual, other proof of the same identity. I just like it because it seems to be very elegant. Obviously, it was historically found or offered, I know, much later than this original one because the original is kind of a natural, simple. It goes exactly about a number of combinations. So, it's expressed in combinatorial terms. The combinatorial identity is expressed in combinatorial terms. Now, I'm going to use pure algebra. Now, 1 plus X to the power of M plus N is equal to 1 plus X to the power of M times 1 plus X to the power of N. Now, we will start from this obvious identity. I will also use the binomial, Newton's binomial of what is A plus B to the power of N. This is, as you remember, A to the power N, B to the power of 0 plus C from N. But I should actually put this coefficient before this as well. So, it's C and another thing is I probably shouldn't use N because N I'm using for other purposes. So, let's put K. K 0 A K B 0 plus C K 1 A K minus 1 B 1, etc. And the last couple of numbers, K minus 1 A to the power of 1 B to the K minus 1 plus C K K A 0 B K. Okay, this is Newton's binomial. And again, I hope that we all know this. If we don't, there is a special section in this course which basically proves this. Anyway, so I'm using this binomial formula for left and the right side, right? And I will use the sigma notation. Let me just replace this with sigma notation. It would be much more convenient for me. So, it's sigma I from 0 to K C K I A to the power of I B to the power of K minus I. Alright, so this is sigma I from 0 to M plus N C M plus N I times A is 1 and B is X. So, 1 to any power is 1. So, we're talking about only B which is X. So, it will be X to the power of M plus N minus I. Okay, equals 2. Here, I will do exactly the same with each one of them. The first one is sigma I from 0 to M. The number of combinations by I X to the power of M minus I times 1. Sorry, sigma I from 0 to N C N I and X to the power of N minus I. Okay. Now, let's just think about it. These are two polynomials which depend on X. The power of this polynomial, the maximum X is when I is equal to 0 which is M plus N. The power of this is M and the power of this is N and the product of two polynomials of M and N's power obviously is the polynomial of M plus N's power. So, that actually corresponds. Well, but not only that. Basically, each coefficient of this polynomial must be equal to the corresponding coefficient with X to the same power of this polynomial, right? Because it's true for any X. So, if two polynomials are equal to each other for any X, that means that each coefficient of these polynomials must be exactly the same or else. All right. So, let's just do the coefficient at one particular X namely X to the power of R. Now, what would be the coefficient X to the power of R on the left? Well, it would be, if this is equal to R, then I is equal to... Well, you know what? It would be probably much easier for me if I would do it slightly differently. You see, this is exactly the same but it starts with 0 which means this should start with 0. It's completely symmetrical. So, I will start with I here and I here and I here. Because what I wrote before was actually X plus 1, not 1 plus X. But it's exactly the same thing because the coefficients, the number of combinations is exactly symmetrical. So, M plus N minus I is the same as I. So, it's completely symmetrical but this would be easier for me to prove. Because now the coefficient at X to the power of R on the left is C M plus N by R, number of combinations. Now, let's think about what is the coefficient at X to the power of R on the right side? Well, this is a product of two sums. Now, this is a polynomial where the power goes from 1 to M and this is polynomial when the power goes from 1 to M. Now, whenever I have two polynomials, I'm basically having all the different possible products. So, if I have, let's say, certain number of members here and certain number of members here, then the total number of these elements would be the product. So, if I would like to have the power X to the power of R, I can take the element with X to the power of 0 here and X to the power of R there. Or I can take an element X to the power of 1 here and X to the power of R minus 1 here, etc. So, basically what I'm saying is that out of all the different combinations of coefficients this times this, I need only those coefficients which have a sum of powers equal to R, which is the coefficient at X0, which is Cm0 times coefficient of this by X to the power of R, plus coefficient at X to the first power times coefficient of X to the power R minus 1, etc. So, basically as you see, we will get, and the last one would be this, this. And they must be equal. So, which is basically another proof of the same Van der Monde's identity using the Newton's binomial formula, which is basically algebra. Anyway, personally I do like the first proof much better because it's very tangible, so to speak. This, you have to go back to the binomial formula for A plus B to some power, you have to apply it, and then you have to really think how the sum of different members are multiplied to each other, that it's all the different pairs of one member from here and one member from there. And these are sum of all members which give X to the power of R, and that's why we have this identity. Well, that's basically it. It's relatively simple thing, two different proofs of the same fancy named Van der Monde's identity. I do recommend you to go to the website and check the notes for this lecture. They are rather detailed, basically the same thing which I did here, but it'll be just one more repetition. That's it, thank you very much and good luck.