Answer
See below.
Work Step by Step
If we want to choose $k$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $_{n}C_k=\frac{n!}{(n-k)!k!}$ ways.
The order doesn't matter here in the set of players, thus we have to use combinations. Thus $_{4}C_{1}(the \ pitcher)\cdot_{15-4}C_{9-1}(the \ remaining \ players)=165\cdot4=660$
