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January 31, 2015

January 31, 2015

Posted by **Manny** on Friday, June 22, 2012 at 2:20pm.

- algebra -
**MathMate**, Friday, June 22, 2012 at 3:15pmAssuming 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!]

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