Prove by a combinatorial argument that ${n \choose r}{r \choose s}={n \choose s} {n-s \choose r-s} $

Question

Prove by a combinatorial argument that ${n \choose r}{r \choose s}={n \choose s} {n-s \choose r-s} $

Is a little hard for me solve this problem.

I see we need to use the multiplication principle. But is a little hard to me finding the idea for prove this...

Can someone give me a hint?

Answer 1

Choose an $r$-subset of $n$ and from this subset choose an $s$-subset. This is the same as choosing an $s$ subset of $n$ and from the complement of size $n-s$ an $r-s$ subset. Hence the result.

Answer 2

We can count the number of ways of forming a committee of $r$ people with a subcommittee of $s$ people from a group of $n$ people in two ways.

1. Choose the members of the committee, then select the members of the subcommittee.
2. Choose the members of the subcommittee, then select the remaining members of the committee.

The left-hand side counts the number of ways of selecting a committee of $r$ people from a group of $n$ people and then selecting $s$ of those $r$ people to serve on the subcommittee. The right-hand side counts the number of ways of selecting a subcommittee of $s$ people from a group of $n$ people and then selecting the remaining $r - s$ members of the committee from the remaining $n - s$ people in the group.

Answer 3

For sake of variety, here's an algebraic proof:

$$\begin{align*} {n \choose r}{r \choose s} &=\frac{n!}{r!(n-r)!}\frac{r!}{s!(r-s)!}\\\\ &=\frac{n!}{(n-r)!\cdot s!\cdot(r-s)!}\\\\ &=\frac{n!}{s!\color{red}{(n-s)!}}\frac{\color{red}{(n-s)!}}{(r-s)!(n-r)!}\\\\ &=\frac{n!}{s!(n-s)!}\frac{(n-s)!}{(r-s)!(n-s-(r-s))!}\\\\ &={n \choose s}{n-s \choose r-s} \end{align*}$$