There are 12 copies of Beetles CDs, 7 copies of Pink Floyd CDs, 3 different CDs of Michael Jackson, and 2 different CDs of Madonna. Find the number of ways in which one or more than one CD can be selected?

Copies of Beetle CD's = 12
Copies of Pink Floyd CD's  = 7
Different CD's of MJ = 3
Different DC's of Madonna = 2 

In the case of 5 (3+2) different CD's of MJ & Madonna, each CD has two possibilities that is either be selected or not selected . Hence for different CD's it would be $$2^3\times2^2$$ ways.

In case of copies of Beetle CD's either 0 can be selected or 1 can be selected or 2 can be selected & so on.. till 12 i.e {0,1,2,3...12} . Hence for this no of ways will be 1+1+1+1....13 times = 13 ways.

In case of copies of Pink Floyd CD's either 0 can be selected or 1 can be selected or 2 can be selected & so on.. till 7 i.e {0,1,2,3...7} . Hence for this no of ways will be 1+1+1+1....8 times = 8 ways

Total no of ways of selection = $$2^5$$x13x8 = 3328. 
In the no of ways calculated above, there is one way in which none of them are selected , but the question is about atleast 1 or more to be selected , Hence subtract 1 from above number i.e 3327 is no of ways in which one or more can be selected.

Therefore, the answer is 3327.