Alternating sign Vandermonde convolution
Solution 1:
$$(1-x)^r(1+x)^s=\left(\sum_{g=0}^r (-x)^g{r\choose g}\right)\left(\sum_{h=0}^sx^h{s\choose h}\right)$$
$$\implies \sum_{k=0}^n(-1)^k{r\choose k}{s\choose n-k}=[x^n](1-x)^r(1+x)^s.$$
How closed would you consider this? I'm not sure if it gets simpler, but obviously it tells us
$$\sum_{k=0}^n(-1)^k{r\choose k}{r\choose n-k}=\begin{cases}0& n\text{ odd}\\ \\ {r\choose n/2}& n\text{ even}\end{cases}.$$
Solution 2:
According to Maple, the answer is ${s\choose n}{{}_2F_1(-r,-n;\,s-n+1;\,-1)}$ (of course we must assume $s \ge n$ for this to make sense).