Since the problem does not take the order into account this is a combination problem.
A combination is given by:
[tex]_nC_k=\frac{n!}{k!(n-k)!}[/tex]Then in this case we have:
[tex]_{14}C_5=\frac{14!}{5!(14-5)!}=2002[/tex]Therefore there are 2002 different collections of five books
