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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?

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

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

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