<s>[INST] Prove, using a combinatorial argument, that $\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$ 
Prove, using a combinatorial argument, that
$$\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$$

attempt: In $m$, suppose that $n$ went for a ride, and out of those $n$, $r$ were awarded. This can be done in $\binom{m}{n}\cdot\binom{n}{r}$ ways. Now suppose we choose $r$ people to be rewarded from these $m$, and then we choose the $n-r$ people who only went on the tour but were not rewarded: $\binom{m}{r}\cdot \binom{m-r}{n-r}$ . So $$\binom{m}{n}\cdot\binom{n}{r}= \binom{m}{r}\cdot \binom{m-r}{n-r}$$
I'm right?
 [/INST] Solution:
For the left part, we have
$$
\left( \begin{array}{c}
 m\\
 n\\
\end{array} \right) \cdot \left( \begin{array}{c}
 n\\
 r\\
\end{array} \right) =\frac{m!}{n!\left( m-n \right) !}\cdot \frac{n!}{r!\left( n-r \right) !}=\frac{m!}{\left( m-n \right) !\left( n-r \right) !r!},
$$
and for the right part, we have
$$
\left( \begin{array}{c}
 m\\
 r\\
\end{array} \right) \cdot \left( \begin{array}{c}
 m-r\\
 n-r\\
\end{array} \right) =\frac{m!}{r!\left( m-r \right) !}\cdot \frac{\left( m-r \right) !}{\left( n-r \right) !\left( m-n \right) !}=\frac{m!}{\left( m-n \right) !\left( n-r \right) !r!}.
$$
Therefore, we have
$$
\left( \begin{array}{c}
 m\\
 n\\
\end{array} \right) \cdot \left( \begin{array}{c}
 n\\
 r\\
\end{array} \right) =\left( \begin{array}{c}
 m\\
 r\\
\end{array} \right) \cdot \left( \begin{array}{c}
 m-r\\
 n-r\\
\end{array} \right).
$$
</s>