#### Answer

$C(n,r)=\dfrac{n!}{(n-r)r!}$

#### Work Step by Step

We know that $P(n,r)=n(n-1)(n-2)...(n-r+1)=\dfrac{n!}{(n-r)!}.$
But in combination, the order of $r$ objects doesn't matter, hence we divide $P(n, r)$ by $r!$ (because $r$ objects can be arranged in $r!$ ways).
Hence,
$C(n,r)=\dfrac{n!}{(n-r)r!}$