Show $\binom{a}{b}\binom{b}{c}=\binom{a}{a-c}\binom{a-c}{a-b}$

Question

I am trying to give a non-algebraic proof for this equality:

$$\dbinom{a}{b}\dbinom{b}{c}=\dbinom{a}{a-c}\dbinom{a-c}{a-b}$$

So far, I could only use the identity $\dbinom{x}{y}=\dbinom{x}{x-y}$.

What I got is $\dbinom{a}{a-c}\dbinom{a-c}{a-b}=\dbinom{a}{c}\dbinom{a-c}{b-c}$. But I don't think I have made any progress from there. It seems reasonable if we think about the identity, but I just find it difficult to give a proper proof for it.

Thanks for the help.

Answer 1

HINT

Here is the basis for a combinatorial proof.

Try and imagine that there are $a$ people. Of these, $b$ people will join group $A$, and we will pick $c$ people as the committee of group $A$. It is not too difficult to show this is the left-hand side.

For the right-hand side, try picking the committee first, then the remaining members of the group.

The number of ways you pick them should be equal both times.