Show that ${n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}$ Show that $\begin{align}{n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}.\end{align}$
Not sure how to approach this exactly. I've tried to use the property $\begin{align}{n \choose k} = {n \choose n-k} \end{align}$, which seems like it could be useful, but doesn't gett me anywhere.

Hint: $$\binom{a}{b} = \frac{a!}{b!(a-b)!}.$$
