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:
$$={t+1 \choose t-r-s} = {t+1 \choose r+s+1}.$$
Let $r$, $t$, $s$ be fixed.

$\binom{t+1}{r+s+1}$ = number of possibilities how can I choose $r+s+1$ elements from $\{1,2,\dots,t+1\}$

Let us order the chosen elements increasingly: $a\_1 < a\_2 < \dots < a\_{r+s+1}$

What is the number of possibilities where $a\_{s+1}=k+1$? We have to choose $s$ elements from $\{1,2,\dots,k\}$ and the remaining $r$ elements from $\{k+2,\dots,t+1\}$. We have
$$\binom ks \cdot \binom {t-k}r$$
possibilities. 

The last expression is non-zero only for $k\ge 0$ and $t-k\ge 0$, which gives us the range of summation.

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Although you are probably not interested in a combinatorial proof, since you explicitly mentioned generating functions.