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.

Use the fact that (which follows from the definition)
$$\binom{m}{k}=\binom{m}{m-k}.$$
Once you have this the LHS can be written as 
$$LHS=\sum_{k=0}^n\binom{n}{k}\binom{m}{k}=\sum_{k=0}^n\binom{n}{k}\binom{m}{m-k}.$$
Now we can do a combinatorial argument to find this sum. Consider a group of $n$ men and $m$ women. We want to make a committee consisting of $m$ people. This can be done in any of the following ways:
1). $0$ men and $m$ women-----this selection can be made in $\binom{n}{0}\binom{m}{m}$ ways.
2). $1$ man and $m-1$ women-----this selection can be made in $\binom{n}{1}\binom{m}{m-1}$ ways.
and so on.....
m+1). $m$ men and $0$ women-----this selection can be made in $\binom{n}{m}\binom{m}{0}$ ways.
The sum total of this gives you the LHS. But this problem can also be solved by considering choosing $m$ people out of a group of $m+n$ people, which can be done in $\binom{n+m}{m}$ ways. Hence the two ways of counting should be equal. 
$$\sum_{k=0}^n\binom{n}{k}\binom{m}{m-k}=\binom{n+m}{n}=\binom{n+m}{m}$$ 
