# Proving a combinatorial identity involving sum of products of binomial coefficients

I am trying to prove the following identity:

$$\sum_{n=k}^{s-r} \binom{n}{k} \binom{s-n}{r} = \binom{s + 1}{k + r + 1}$$

It looks kind of like a flipped version of Vandermonde's identity but I haven't been able to make any progress in proving it. I've verified it for random small values of $$k, r$$ and $$s$$ using a computer.

Suppose that you have a row $$s+1$$ balls, and you want to choose $$k+r+1$$ of them; clearly that can be done in $$\binom{s+1}{k+r+1}$$ ways.
Now we’ll break that down according to the position of the $$(k+1)$$-st of the chosen balls. There must be at least $$k$$ balls to the left of it, so the first position in which it could possibly occur is position $$k+1$$. There must be at least $$r$$ balls to right of it, so the last position in which it could occur is position $$s+1-r$$. Thus, number of balls to the left of it must be at least $$k$$ and at most $$s-r$$.
Let $$n$$ be the number of balls to the left of it, so that $$n$$ ranges from $$k$$ through $$s-r$$. The $$k$$ chosen balls that precede it can be any $$k$$ of the $$n$$ balls to its left, so there are $$\binom{n}k$$ ways to choose them. The $$r$$ chosen balls that follow it can be any of the $$s-n$$ balls to its right, so there are $$\binom{s-n}r$$ of them. Thus, there are $$\binom{n}k\binom{s-n}r$$ ways to choose the $$k+r+1$$ balls so that there are $$n$$ balls to the left of the $$(k+1)$$-st chosen ball. Summing over the possible values of $$n$$ will give us the total number of ways to choose $$k+r+1$$ balls, so
$$\sum_{n=k}^{s-r}\binom{n}k\binom{s-n}r=\binom{s+1}{k+r+1}\;.$$