combinatorics - Prove by a combinatorial argument that $(n-r){n choose r}=n{n-1 choose r}$

Prove by a combinatorial argument $$(n-r){n \choose r}=n{n-1 \choose r}$$

My attempt:

We have two ways of count the number of persons forms a committee of a group $n$ of people.

Here I'm a little confused, because I don't know how interpret the multiplication by $(n-r)$ here. Can someone help me?

Answer

On the RHS

• we choose one president ($n$ choiches) and then form a committee of $r$ out of n-1

On the LHS

• we form a committee of $r$ out of $n$ and then choose a president from the others $n-r$