Evaluate $\sum_{k=0}^{n} {n \choose k}{m \choose k}$ for a given $n$ and $m$. How do I evaluate $\sum_{k=0}^{n} {n \choose k}{m \choose k}$ for a given $n$ and $m$.
I have tried to use binomial expansion and combine factorials, but I have gotten nowhere. I don't really know how to start this problem.
The answer is ${n+m \choose n}$. Any help is greatly appreciated.
EDIT: I'm looking for  a proof of this identity.

Suppose we want to pick $n$ children from a group of $n$ boys and $m$ girls. Then we can pick $n$ boys and $0$ girls, or $n - 1$ boys and $1 $ girl, or $n - 2$ boys and $2$ girls, ... There are
$$\sum_{r = 0}^n\binom{n}{n - r}\binom{m}{r} = \sum_{r = 0}^n\binom{n}{r}\binom{m}{r}$$
ways to do this.
But we can also look at it as choosing $n$ objects from a set of $n + m$ objects. Then there are
$$\binom{n + m}{n}$$
ways to do this.
Since both are counting the same number of things, they must be equal, i.e.
$$\sum_{r = 0}^n\binom{n}{r}\binom{m}{r} = \binom{n + m}{n}$$
