Show that ${n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}$

Question

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.

Answer 1

From $X$ with $|X| =n$, choose two subsets $A$ and $B$ such that $A \subset B$ and $|A| = m, |B| = k$

You can firstly choose $B$ from $X$(${n \choose k}$ ways), then choose $A$ from $B$(${k\choose m}$ ways).

Alternatively, you can firstly choose $A$ from $X$(${n \choose m}$ ways), then add $k-m$ anther elements into $A$ to form $B$( ${n-m\choose k-m}$ ways).

Answer 2

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

Answer 3

HINT: You can do it by a straightforward calculation with factorials, using the fact that $\binom{n}k=\frac{n!}{k!(n-k)!}$, or you can do it combinatorially. For the combinatorial reasoning in the same problem in a slightly different guise see this answer.