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

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