Combinatorial argument $a(n-a)$ $n \choose a $ = $n(n-1)$ $n-2 \choose a-1$

Question

I can not make sense of this; I am looking for a combinatorial argument that would prove the equivalence of this statement. I can prove it with algebraic manipulation.

$a(n-a)$ $n \choose a $ = $n(n-1)$ $n-2 \choose a-1$

Thank you

Answer 1

Choose $a$ people from $n$ to form a committee; choose one member of the committee to be president, and choose someone not on the committee to be an independent observer.

Method 1: choose the committee in $C(n,a)$ ways. Choose the president, $a$ ways. Choose the observer, $n-a$ ways.

Method 2: choose the president, $n$ ways. Choose the observer, $n-1$ ways. Choose the other $a-1$ committee members from the remaining $n-2$ people, $C(n-2,a-1)$ ways.

Therefore $$a(n-a)\binom{n}{a} = n(n-1)\binom{n-2}{a-1}\ .$$

Answer 2

I think you meant $$a(n-a)\binom{n}{a} = n(n-1)\binom{n-2}{a-1}.$$

Think about choosing $a+1$ people out of $n$ people so that there is a group leader and co-leader. You can first choose $a$ people, choose the leader among those $a$ people and the co-leader in the rest $n-a$. Or you can first choose the leader, then the co-leader, and then the rest $a-1$ people.