In this situation, the order matters, since being 1st pick is not the same as being 2nd or 3rd. Therefore, there are 14 choices for the first pick, 13 remaining for the second pick, and 12 for the third pick. This is equivalent to the permutation 12P3.
[tex]12P3 = \frac{12!}{(12-3)!} = 12(11)(10) = 1320[/tex] ways
(If order did not matter, this would use a combination.)
