Prove by a combinatorial argument that ${n \choose r}{r \choose s}={n \choose s} {n-s \choose r-s} $ 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?
 A: For sake of variety, here's an algebraic proof:
$$\begin{align*}
{n \choose r}{r \choose s}
&=\frac{n!}{r!(n-r)!}\frac{r!}{s!(r-s)!}\\\\
&=\frac{n!}{(n-r)!\cdot s!\cdot(r-s)!}\\\\
&=\frac{n!}{s!\color{red}{(n-s)!}}\frac{\color{red}{(n-s)!}}{(r-s)!(n-r)!}\\\\
&=\frac{n!}{s!(n-s)!}\frac{(n-s)!}{(r-s)!(n-s-(r-s))!}\\\\
&={n \choose s}{n-s \choose r-s}
\end{align*}$$
A: 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. 
A: 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.


*

*Choose the members of the committee, then select the members of the subcommittee.

*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. 

