Question: 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.

Answer: 
I'm going to make more explicit the point I think Phira is making. The identity really is just Vandermonde's convolution plus the upper negation rule for binomial coefficients. 

The upper negation rule for binomial coefficients is $$\binom{n}{k} = (-1)^k \binom{k-n-1}{k},$$ which holds when $k$ is an integer (see, for example, *Concrete Mathematics*, 2nd edition, p. 164).
Applying this, we get
$$\sum\_{0 \le k \le t} {t-k \choose r}{k \choose s} = \sum\_{0 \le k \le t} {t-k \choose t-k-r}{k \choose k-s}= \sum\_{0 \le k \le t} (-1)^{t-k-r}{-r-1 \choose t-r-k} (-1)^{k-s}{-s-1 \choose k-s}$$
$$ = (-1)^{t-r-s}\sum\_{0 \le k \le t}{-r-1 \choose t-r-k}{-s-1 \choose -s+k}.$$
Now, use Vandermonde's convolution (or, more generally, the [Chu-Vandermonde identity](http://en.wikipedia.org/wiki/Vandermonde%27s_identity#Chu.E2.80.93Vandermonde_identity)),
to get 
$$ = (-1)^{t-r-s}{-r-s-2 \choose t-r-s},$$
and then apply the upper negation rule again, which gives us what we want:
$$={t+1 \choose t-r-s} = {t+1 \choose r+s+1}.$$
Note that this summation *is* Vandermonde's identity.

Calculate in both sums the ratio of consecutive summands and compare them. You will see that after a suitable change of variables they are the same. 

Therefore, the two sums only differ by a global factor in each term and the result.

If you want to know more about this, read about [hypergeometric functions](http://en.wikipedia.org/wiki/Generalized_hypergeometric_function).