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

Question

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?

Answer 1

Maple gets $$ {\alpha\choose n}{\mbox{$_2$F$_1$}(-n,-\beta;\,\alpha-n+1;\,-1)}$$