Proof of  $\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1}$? How do I prove that
$$\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1} \>?$$
I saw this in a book discussing generating functions.

Suppose we seek to verify that
$$\sum_{0\le k\le t} {t-k\choose r} {k\choose s}
= {t+1\choose r+s+1}.$$
Introduce
$${t-k\choose r} = {t-k\choose t-k-r} =
\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-k-r+1}} (1+z)^{t-k} \; dz.$$
This controls the range so we may extend $k$ to infinity,
getting for the sum
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-r+1}} (1+z)^{t} 
\sum_{k\ge 0} {k\choose s} \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-r+1}} (1+z)^{t} 
\sum_{k\ge s} {k\choose s} \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-r+1}} (1+z)^{t} \frac{z^s}{(1+z)^s}
\sum_{k\ge 0} {k+s\choose s} \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-r+1}} (1+z)^{t} \frac{z^s}{(1+z)^s}
\frac{1}{(1-z/(1+z))^{s+1}} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-r-s+1}} (1+z)^{t+1} 
\frac{1}{(1+z-z)^{s+1}} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\frac{1}{z^{t-r-s+1}} (1+z)^{t+1} 
\; dz.$$
This evaluates to
$${t+1\choose t-r-s} = {t+1\choose r+s+1}$$
by inspection.
