Let $0\le c \le b \le a $. Show the identity is true with combinatorial proof. $ {a \choose b} {b \choose c} = {a \choose a-c}{a-c \choose b-c}$. By using algebra, I can find out that both sides are dividing $a$ into 3 sets $(a-b), (b-c), c$, but I have no clue how to use combinatorial proof. The multiplication part is confusing.
Edit: I didn't know this question has been asked before. After looking at the hint, is it correct to say we choose committee first because we have ${a \choose a-c}$ which is same as ${a \choose c}$?
Edit: LHS is saying we choose $b$ out of $a$ into group I and then we choose $c$ out of $b$ into group II. RHS is saying we take $a-c$ out of $a$ in the group II. Then group II would be left with $c$. Next we choose $b-c$ out of $a-c$ into group I. Thus, group I and group II have same numbers on both sides. Is this right? And is this a complete proof?