Given that a, b, c ∈ N and a ≥ b ≥ c, prove that $\binom{a}{c}\binom{a-c}{b-c}=\binom{a}{b}\binom{b}{c}$.
To start, we know that $\binom{a}{c} = \frac{a!}{c!(a-c!)}=\frac{a!}{(a-c!)c!}=\binom{a}{a-c}$.
Then we have to get $\binom{a-c}{b-c}$.
But when I reach $\binom{a-c}{b-c} = \frac{a-c!}{(b-c)!((a-c)-(b-c)!)}=\frac{a-c!}{(a-b)!(b-c)!}$, I don't know how to proceed.
How do I continue solving this?
The left side is $\frac {a!} {{c!}{(a-c)!}} \frac {(a-c)!} {{(a-b)!}{(b-c)!}}$. Right side is $\frac {a!} {{b!}{(a-b)!}} \frac {b!} {{(b-c)!}{c!}}$. Can you see that these two are equal?