[MUSIC] So, we have seen the inductive rule for coefficients. But what about the explicit formula? Well, for usual binomial coefficients, the formula is well known, (m + n, m ) = ( m + n )! Divided by m! Times n! It turns out that literally the same analogue over this formula holds for q binomial coefficients but to state it we first need to state the right notion of integer numbers and factorials. Namely what are acute analogues of integers and what is the q analog of n factorial? So, q-integer numbers. So what is an integer number? Well, we'll denote it by n in squared brackets and we define it as a strange way, so n is the number of one element subsets of an N elements set. So this is, n choose one and this is my definition. And this is one plus q plus q squared plus etc, and we sum up until q to the power n minus one. Note that here, we sum up until n minus one, not until q to the power n. So this is the generating functions for young diagrams inside the rectangle of size 1 times n- 1. Okay, and this can be written down as fraction, 1-q to the n divided by 1-q. So this will be called the q integer number by definition. So. So our general rule for evaluating things at q equal to 1, halt. So if we find a value of this binomial in q equal to 1, this will be exactly n in the usual sense. So, n evaluated at q equals to 1, is just n. Okay. How to sum up integers. So if we add n in square brackets to m in square brackets. This, of course this won't be equal to m + n in squared brackets. Well, these two polynomials are different. They have different degrees and so on. But fortunately, an analog of this formula holds. So if we, Multiply one of these numbers by q to the power of the first number. So if we multiply this n is squared brackets by q to the power of n or alternatively if we take n and add it with q to the power of n times m, what we get is exactly the QL of m plus n. We will need this formula and the following part. Okay. So what about factorials? We define factorials in obvious way so if n factorial is 1 times 2 times etc times N, then what is the analog of, factorial of Q not Q integer N. So this will be by definition, Q analog of 1 times. Q analof of 2 times etc times N in squared brackets. And also by definition We define zero factorial to be equal to one. Okay, so now comes our proposition, the analog of this quality. M + N, choose m. Equals m + n factorial. Divided by m factorial. [BLANK AUDIO] Times n factorial. So we need to replace all the integers and this formula by their q analogues. The usual binomial efficient by its q-analogue and the same formula will. So if you eliminated as Q equal to one you will get exactly the same equality. Okay, let's prove it. We are proving by induction or m + n If m + n = 1. Then our quantity is obvious. So here's the induction step. Suppose our equality holds for m + n- 1 Now the left hand side of the equality which is m+n choose m is equal to m+n-1 choose m-1 + q to the power of m times m+n-1 choose m. And, My induction hypothesis, this is equal to m plus n minus one factorial, over m minus one factorial times n factorial. We are dealing with q-analogs here. Plus q to the m, m+n-1 factorial over m factorial times n-1 factorial. So, of these two fraction have common multiple. [m + n- 1] factorial over [m- 1] factorial times [n- 1] factorial. And it is multiplied by 1 over n in squared brackets plus q to the power m divided by m in squared brackets. And this is equal to. M + n- 1 factorial, over m- 1 factorial times n- 1 factorial, times the sum, Which is equal to [m] + q to the power m, Times n, Divided by the common denominator. M times n. And using this observation. We see that the numerator here is nothing but m plus n. Which gives us the desired equality. That this is equal to m+n factorial over m factorail times n factorial. The proposition's proved. [MUSIC]