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.

These are just two different ways to express the trinomial coefficient $\binom n{a,b-a,n-b}$ as a product of two binomial coefficients. The relevant formula (not obviously present on the wikipedia page) is
$$
   \binom n{k,l,m} = \binom nk\binom {n-k}l \qquad\text{whenever $k+l+m=n$,}
$$
together with the symmetry with respect to $x,y,z$ of the trinomial coefficient, and similarly for binomial coefficients (so that $\binom nb=\binom n{n-b}$). The formula above is a consequence of $(X+Y+Z)^n=(X+(Y+Z))^n$, and similar formulas hold for higher multinomial coefficients.
