[MUSIC] Okay, this is side bar. Let me tell you about another appearance of q-binomial coefficients in linear algebra over finite fields. So, if you don't know what a finite field is you can just skip this piece. Suppose that q is a power of prime. Q is p to the power, let's say, capital N. And then, we can consider a finite field with q elements. This field will be denoted by Fq. >> Okay, consider a vector space over Fq of dimension m+n. V is isomorphic to Fq to the power m+n. We can consider n-dimensional subspaces inside V. Okay, Fq is a finite field so there are finite elements such as subspaces. We can ask ourselves how many of them are there? So, the answer is that this number is exactly the q-binomial coefficient, m+n, choose n. Theorem, The # { subspaces, Inside Fq to the power m+n of dimension =n} = Q-binomial coefficient [m+n, choose n]. Let's prove this. Okay, Let us pick a basis inside U, let, In how many ways can we do that? In how many ways can we pick n linearly independent vectors inside of V? So to pick u1, we have Q to the power m+n -1 possibilities. Then, u2 well, -1 here is because the first base is vector should be non-zero but a part of the zero vector it can be an arbitrary vector of V. Okay, then to pick u2, we have q to the power m+n- q possibilities. Because we want our vector u2 be linearly independent from u1 and so it should be not contained in the line span by u1. And the line span by u1 contains exactly q vector. So, u2 should not be contained in the span of u1. And this line has cardinality q. So for u3, There are q to the power m+n- q squared possibilities. Because u3, Should not be contained inside the planes and by u1 and u2. And this plane consists of q squared points. For the last vector, un, the number of possibilities to pick it is q to the power m+n- q to the power n-1. Because the condition is that, it should not be contained in the subspace panned by the previous vectors, u1 et cetera un-1. So, the total number of tuples of n linearly independent vectors is equal to (q to power of m+n- 1).(q to the power m+n- q).et cetera.(q to the power m+n- q to the power n-1). And this is the number of tuples. (u1 et cetera un) which are linearly independent. Each tuple, spans and n-dimensional space U. But given subspace U, Correspond several tuples. Let us compute their number. So, for a given, U says tuple will be basis of U, so then, number of bases, Can be found in a similar way. It is, (q to the power n- 1).(q to the power n- q).et cetera.(q to the power n- q to the power n-1). So, the number of subspaces equals the ratio of this amount, the number of tuples, divided by the number of tuples defining each given U. So, #{U inside Fq to the power of m+n of dimU=n} = (q to the power m+n- 1).et cetera.(q to the power m+n- q to the power n-1)/q to the power n- 1)et cetera(q to the power n- q to the power n-1). And it is obvious that this thing equals The q-binomial coefficient, [m+n choose n]. [MUSIC]