# Is there a closed formula for $\sum_{k=0}^n \binom{\alpha}{n-k} \binom{\beta}{k}(-1)^k$?

Suppose $$\alpha, \beta, n$$ are three non-negative integers with $$n\leq \min(\alpha, \beta)$$. Is there a closed formula (or a combinatorical concept) for $$C_n^{(\alpha, \beta)}:=\sum_{k=0}^n \binom{n}{k} (-1)^k (\alpha)_{n-k}(\beta)_{k}=n! \sum_{k=0}^n \binom{\alpha}{n-k} \binom{\beta}{k}(-1)^k$$ here $$(x)_k=x(x-1)\cdots(x-k+1)$$ is the falling factorial. If that $$(-1)^k$$ was not there then the formula would've been $$(\alpha+\beta)_n$$. Also, is there a combinatorical intepretation of this number?

Maple gets $${\alpha\choose n}{\mbox{_2F_1}(-n,-\beta;\,\alpha-n+1;\,-1)}$$
• What does the notation ₂F₁ mean? – Gregory Nisbet Dec 19 '18 at 0:59