[MUSIC] Okay, now, we're going to sketch another proof of the same theorem, that the number of partitions of n into odd summands is equal to the number of partitions into distinct summands. And this proof will be purely commentarial. But I will not give a complete proof, I will just give a construction and leave it to you as an exercise. Okay, so here's the sketch of the proof. We will construct a bijection between partitions into odd summands and partitions with distinct summands. The best way to explain this construction is to show how it works on an example. And let me tell you that this bijection is due to Sylvester Consider a partition of a number into odd summands. Let's say I will take the partition of 23 as 7 + 5 + 5 + 3 + 1 + 1 + 1. Okay, and now let me draw a Young diagram corresponding to this partition but let me draw it in quite an exotic way. These summands are odd and let me align rows of lengths 7, 5, 5 etc. Not by the left column but by the center. So I take the row with seven boxes one, two, three, four, five, six and seven. Underneath I draw a row with five boxes. One, two, three, four, five. And these rows are both center-adjusted. Then I draw another five boxes. Then three. And then three rows of one box each. Okay, so, I get the following symmetric picture. Now having this picture, I will produce a presentation of the same number into distinct summands. How I'm going to do this? Let me tell you. I will decompose my diagram into hooks. And I will start from the center, I will take the central column. And then I will take the hook formed by the central column and all the boxes from the right of the top box in this column. So I get such a hook of length one, two, three, four, five, six, seven, eight, nine, ten. Okay, so I write 23 is equal to 10 +, let me now draw the second hook. I start from the column, Just directly to the left of the central one and I take the hook formed by these boxes. So I have one, two, three, four, five, six of them. So I write 10 + 6 Okay, and now comes the next hook. It is formed by the boxes under the first one. One, two, three, four. So here's the third one. So we get 4 as the next summand in my decomposition. And what I'm left with is just this head of two hooks. The first one consists of only two boxes, and the last one is just one box here. So I get 2 and 1. So note that each of the boxes is containing one of these hooks. So I get a partition of the same number 23, the number of boxes in my Young diagram. Well, and it turns out, I'll leave it to you as an exercise to show that all the numbers in this decomposition are pair-wise distinct. So all hook lengths are distinct. So, this is a partition into distinct summands. And I started with a partition into odd summands. So now, we need to prove that this is indeed a bijection, that every partition into distinct summands can be produced in such a way from unique partition into odd summands. And this is an exercise for you. Show that this is indeed a bijection. [MUSIC]