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.

Using the Vandermonde Identity, this is
$$
\sum_{k=0}^n\binom{n}{n-k}\binom{m}{k}=\binom{n+m}{n}
$$

Proof of Vandermonde's Identity
$$
\begin{align}
\sum_{k=0}^{m+n}\color{#C00000}{\binom{m+n}{k}}x^k
&=(1+x)^{m+n}\tag{1}\\
&=(1+x)^m(1+x)^n\tag{2}\\
&=\sum_{j=0}^m\binom{m}{j}x^j\sum_{k=0}^n\binom{n}{k}x^k\tag{3}\\
&=\sum_{j=0}^m\sum_{k=j}^{n+j}\binom{m}{j}\binom{n}{k-j}x^k\tag{4}\\
&=\sum_{k=0}^{m+n}\color{#C00000}{\sum_{j=0}^k\binom{m}{j}\binom{n}{k-j}}x^k\tag{5}
\end{align}
$$
Explanation:
$(1)$: binomial theorem
$(2)$: property of exponents
$(3)$: binomial theorem
$(4)$: substitute $k\mapsto k-j$
$(5)$: change order of summation
Compare the coefficients of $x^k$.
