## Precalculus (10th Edition)

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