Answer
$660$
Work Step by Step
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: $C(n,r)C(m,s).$ (Similarly for more than $2$ groups.)
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 because out of the $4$ pitchers $1$ is chosen and from the rest $11$, $9-1=8$ are chosen$C(4,1)C(11,8)=\frac{4}{1}C(11,3)=4\cdot\frac{11\cdot10\cdot9}{3\cdot2\cdot1}=4\cdot165=660$
