# Proof using a binomial coefficient

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?

• Combinatorial arguments often work best in cases like these when the algebra is just an absolute chore. – Eevee Trainer Apr 16 '19 at 8:25
• Among $a$ students, how many ways are there to form a cricket team with $b$ players among whom $c$ players are bowlers? – Anubhab Ghosal Apr 16 '19 at 8:36

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?

• Yes, so is simple fraction simplification enough to prove that the claim is true? – sdds Apr 16 '19 at 8:42
• Rigth. Just writing in terms of factorials and making some cancellations. – Kavi Rama Murthy Apr 16 '19 at 8:45