Give a combinatorial argument to evaluate sum $\sum_{k}^{n}\binom{n }{k}\binom{m}{k}$

Question

$$\sum_{k}^{n}\binom{n }{k}\binom{m}{k}$$

I'm not really sure how a combinatorial argument would help me evaluate this? I was thinking maybe I'm adding up all the ways that you could divide a group of n+m people into 2 groups of size k?

I'm tempted to just plug in the fact that $$\binom{n}{k} =\frac{n!}{k!(n-k)!}$$but that's not really a combinatorial argument, right?

Answer 1

Choose $k$ things from $n$ and choose $m-k$ things from $m$ ($k$ ranges over $0$ to $\min(n,m)$) ... This is the same as choosing $m$ things from $n+m$.

Thus \begin{eqnarray*} \sum_{k=0}^{\min(n,m)}\binom{n }{k}\binom{m}{k} = \binom{n+m}{m}. \end{eqnarray*}