Prove $\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$ I want to prove this equation,
$$
\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}
$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please. 
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.

Directly, we find that
\begin{align*}
{n \choose a}{n - a \choose b - a} &= \frac{n!}{(n - a)!a!} \frac{(n - a)!}{(n - a - (b - a))! (b - a)!} \\
&= \frac{n!}{a!} \frac{1}{(n - b)!(b - a)!} \\
&= \frac{n!}{(n - b)! b!} \frac{b!}{(b - a)!a!} \\
&= {n \choose b}{b \choose a}
\end{align*}
The third equality follows by multiplying and dividing by $b!$.
