Variation of Chu-Vandermonde binomial sum [duplicate]

The well-known Vandermonde convolution gives us the closed form $$\sum_{k=0}^n {r\choose k}{s\choose n-k} = {r+s \choose n}$$ For the case $r=s$, it is also known that $$\sum_{k=0}^n (-1)^k {r \choose k} {r \choose n-k} = (-1)^{n/2} {r \choose n/2} \quad [n \mathrm{\ is\ even}]$$ When $r\not= s$, is there a known closed form for the alternating Vandermonde sum? $$\sum_{k=0}^n (-1)^k {r \choose k} {s \choose n-k}$$

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).

Solution 3:

You can use the Zeilberger algorithm to find a recurrence for the sum and then use Petkovsek's algorithm to verify that there are no hypergeometric (i.e. essentially products of factorials) closed formulas for the general case.

So, you cannot expect to do better than the expressions proposed in the other answers.