2. Consider the following equality, for $n \ge k \ge r$:
$$ \binom{n}{k} \binom{k}{r} = \binom{n}{r} \binom{n-r}{k-r}. $$
It can be proved as follows:
$$ \binom{n}{k} \binom{k}{r} = \frac{n!}{k!(n-k)!} \frac{k!}{r!(k-r)!} = \frac{n!}{r!(n-r)!} \frac{(n-r)!}{(k-r)!(n-k)!} = \binom{n}{r} \binom{n-r}{k-r}. $$
Give instead a "combinatorial" proof: Give a set that has $\binom{n}{k} \binom{k}{r}$ elements and show, by counting it a different way, that it has $\binom{n}{r} \binom{n-r}{k-r}$ elements.
