Posted by Manny on Friday, June 22, 2012 at 2:20pm.
Assuming order matters, i.e.
A,B,C,D is counted differently as B,A,C,D.
There are 9 choices for the first place, 8 places for the second, 7 places for the third, and 6 places for the fourth. Applying the multiplication rule, there are 9*8*7*6 ways, or
P(9,4) [permutations]
=9!/(9-4)!
If order does not matter, the above must be divided by 4! combinations among the top four, or
number of ways
=C(9,4)
=9!/[(9-4]!4!]