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 balls, thus we have to use combinations.
a)Thus $_{15}C_{5}=3003$
b)Thus $_{15}C_{3}\cdot_{10}C_{2}=455\cdot45=20475$
c)Thus $_{15}C_{5}+_{15}C_{4}\cdot_{10}C_{1}=3003+1365\cdot10=16653$
