## Precalculus (10th Edition)

$336$.
If we have two groups with $n$ and $m$ objects and we choose $r(r\leq n)$ and $s(s\leq m)$ objects from the two groups where the order doesn't matter then the number of combinations is: $C(n,r)C(m,s).$ We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ Hence here $C(4,2)C(8,3)=\frac{4\cdot3}{2\cdot1}\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=6\cdot56=336$.