Answer
$1140$
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 when choosing the set of CDs, thus we have to use combinations and we choose $3$ sets out of $20$.
Thus the number of possibilities is: $_{20}C_{3}=\frac{20!}{(20-3)!3!}=\frac{20\cdot19\cdot18}{3\cdot2\cdot1}=1140$
