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

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.
