Chu-Vandermonde Identity/Proof 4

Proof
From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \paren{n - k}} {n - k} \dfrac r {r - t k}$:

where $r, s, t \in \R, n \in \Z$.

Setting $t = 0$:
 * $\ds \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$

which is the result required.