[MUSIC] We will be speaking about q-binomial coefficients but, usually binomial coefficients are related to the binomial theorem. What about the q version of the binomial theorem? So now, I'm going to give one of the possible interpretations of the binomial theorem involving q-binomial coefficients. So, here's the usual binomial theorem, that (1+x) to the power of n = to the sum of, (x to the power of k) times the corresponding binomial coefficient. For k from 1 to n. [COUGH] Here's the theorem. (1+xq).(1+xq squared).et cetera.(1+xq to the power n)= the sum of, The q-binomial coefficients [n choose k].q to the power k(k+1)/2.x to the power of k. So, when you evaluate this add q equal to 1, this multiple becomes 1 and you recover the standard binomial theorem. Let us prove this theorem. Take the left hand side. And consider it as a polynomial in x with coefficients, these coefficients are polynomials in q. So, this is the sum of for k from 0 to n of x to the power k times certain polynomial ak(q). I claimed that those ak(q) are generating in functions for partitions of certain kind. Namely, ak(q) is the generating function, For Partitions With exactly k parts. Exactly k distinct parts, distinct summons. Not exceeding n. Why is this so? How can you get x to the power of k from expanding this product? You should take xq to a certain power from exactly k multiples. And in each of this multiples, x comes together with some power of q. And all of those powers are distinct, and do not exceed n. So, these are parts of our partition. So such partitions have the form lambda k Greater than or equal to 1. Strictly < lambda k-1 < etc., < lambda 2 < lambda 1 and lambda 1 does not exceed n. Okay, so all of these parts are distinct and do not exceed n. Let me draw them. Here I have k, let's suppose that here I have my n. Okay, having such a partition, I can, Produce another partition fitting inside a rectangle of size k times n-k, without a restriction on powers being distinct. What should I do? Let me remove one box from the last row. Two boxes from the row before last. Et cetera and I remove k boxes from the first row. So, in this case, I say that I obtain a partition of mus which are now nonnegative. Mu k is less than or equal to mu k-1 Et cetera less than or equal to mu 1, and mu 1 does not exceed n-k, where mu 1 is lambda 1-k, mu 2 = lambda 2-(k-1), et cetera, mu k is lambda k-1. Okay, so, I remove that many boxes from each row. And I get a partition which now looks as follows. In this example, the rows will have lengths 3, 3, 1. Okay, and in such a way I can get any partition satisfying these conditions, namely fitting inside the rectangle of size n-k times k. And these partitions are indexed by the q-binomial coefficient n choose k. So, Mu's have the generating function. And choose k sub q. And what about lambdas? The generating function for such lambdas is obtained from the generating function on mu's by multiplying it by a certain multiple of q. So, what is this power of q? I need to multiply by q to the power 1+2+ et cetera +k. So, lambda's have the generating function. [n choose k].q to the power of 1+2+ et cetera +k. This is exactly the number of boxes that we removed here. And this is [n choose k].q to the power of k(k+1)/2. And this is nothing but ak(q). So, this is the coefficient in the front of x to the power of q in the q-binomial theorem. So, the theorem is proved. This was the last lecture of our course, introduction to enumerative combinatorics. Our course is over. I hope that you enjoyed my lectures. I will be glad to receive your feedback. Thank you and goodbye. [MUSIC]
