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 $_{7}C_{2}\cdot_{3}C_{1}=21\cdot3=63$
b)Thus $_{7}C_{3}=35$
c)Thus $_{3}C_{3}=1$