[MUSIC] Okay, we also can study partitions with various restrictions on their parts. For instance, we can consider partitions of the number n such that, all the summands are odd. Let's say, question one. What is the number of partitions of n into odd summands? What is, We will denote by Po(n), o stands for odd. For example. Let us find Po(7). Seven is partitioned into odd summands itself, or 7 = 5 + 1 + 1 or 3 + 3 + 1, or 3 + 1 + 1 + 1 + 1, or 7 times 1. So you can easily see that, we have listed all the partitions of seven into odd parts. And there are five of them. So Po(7) = 5. Okay. And, now we will write down the generating function for Po(n). But before we do that, we can ask ourselves another question. What is the number of partitions of n, into distinct summands? All the rows of our young diagram, should have distinct lengths. So, what is the number of partitions of n into pairwise distinct summands? It will be denoted by Pd(n). [COUGH] Okay, let us compute Pd(7) again. Well, 7 = 6 + 1, or it is equal to 5 + 2 Or 4 + 3, or 4 + 2 + 1, and that's all. So we get one, two, three, four, five, five again. So this number, five is the same as this one. And it turns out that, this is not a coincidence. So the theorem is that, the number of partitions of n into odd summands, is equal to the number of partitions of n, into pairwise distinct summands. Po(n) = Pd(n) or each n. Now, let us prove this theorem. So we would like to show that, the number of partitions of n into odd summands, is equal to the number of partitions of n into distinct summands. We're going to give one proof this theorem, based on generating functions. And also sketch a purely combinatorial bijective proof. So here comes the first one. We'll show that, the generating functions for these two sequences coincide. Okay. In a similar way as we have seen in the previous propositions, we can show that the generating function for the number of partitions of n, into odd parts. Capital Po(q), which is Po(n) times q to the power n, for n greater than or equal to 0, is equal to the product of infinitely many geometric series, with common ratios q, q to the third, q to the fifth etc. So this is (1 + q + q squared + etc.) times (1 + cubed + q to the sixth + etc.) times (1 + q to the fifth + q to the tenth + etc.). And this thing is equal to, the infinite product of 1 over 1- q to the power 2k- 1, for k from 1 to infinity. So it is 1 over 1- q times 1 over 1- q to the third times 1 over 1- q to the fifth times etc. So this is the generating function for Po(n). What about the generating function for Pd(n)? Pd(q) which is the sum of, Pd(n) times q to the nth, can be found in any similar way. It turns out that, this function is the product of infinitely many factors of the following form. (1 + q) times (1 + q squared) times (1 + q) cubed times etc. The proof is also absolutely the same. Okay. So, all we need to show is, to verify that these two expressions are indeed the same? Okay. To do this, let us multiply Po(q) By some factors, each of which will be equal to one. Namely, Po(q) is 1 over 1- q times something which will be identically equal to one, this will be 1- q squared over 1- q squared, times 1 over 1- q cubed, times 1- q to the fourth over 1-q to the fourth. This is also equal to one, times 1 over 1- q to the fifth, times etc. So, this is equal to the following ratio. In the numerator we have, 1- q squared times 1- q to the fourth times 1- q to the sixth times etc. And in the denominator, we have the same thing as we have in the generating function. 1- q times 1- q squared times 1 minus q cubed times etc. And we see that, each of those factors in the numerator is divisible by, the corresponding factor in the denominator. So, we get. 1- q squared divided by 1- q, gives us 1 + q. 1- q to the fourth divided by 1- q squared gives us, 1 + q squared. 1- q to the sixth divided by 1- q to the third gives us, 1 + q to the third. And so on. So what we get is, exactly the expression for Pd(q), for the generating function of our partitions into distinct summands. So this finishes the proof. [MUSIC]