# Alternating sum of a simple product of binomial coefficients: $\sum_{k=0}^{m} (-1)^k \binom m k \binom n k .$

I would like to evaluate the following alternating sum of products of binomial coefficients: $$\sum_{k=0}^{m} (-1)^k \binom m k \binom n k .$$ I had the idea to use Pascal recursion to re-express $\binom n k$ so that we always have $m$ as the upper index and I have been able to come up with nice expressions for $$\sum_{k=0}^{m} (-1)^k \binom m k \binom m {k-j},$$ ($j=0$ gives the central binomial coefficient as is well known and the others turn out to be shifted away from the centre by $j$ steps, up to a sign).

However we then end up with another alternating sum of products of these solutions with the binomial coefficients that came out of using the recursion. And this sum seems to be even worse to solve, at least combinatorially. Any help appreciated! Thanks in advance.

Let $c_k = (-1)^k \binom{m}{k} \binom{n}{k}$. Then $$\frac{c_{k+1}}{c_k} = - \frac{\binom{m}{k+1}}{\binom{m}{k}} \cdot \frac{\binom{n}{k+1}}{\binom{n}{k}} = - \frac{k-m}{k+1} \cdot \frac{k-n}{k+1}$$ Hence $$c_k = c_0 \prod_{i=0}^{k-1} (-1) \frac{i-m}{i+1} \cdot \frac{i-n}{i+1} = \frac{(-1)^k}{k!} \frac{(-m)_k \cdot (-n)_k}{(1)_k}$$ where $(a)_k$ denotes the Pochhammer symbol. Hence the sum in question can be understood as the Gauss hypergeometric function: $$\sum_{k=0}^m (-1)^k \binom{m}{k} \binom{n}{k} = {}_2F_1\left(-m, -n; 1; -1\right)$$
$$\sum_{k=0}^{m} (-1)^k \binom m k \binom n k = P_{m}^{(0,-n-m-1)}(3) .$$