Probability problem (counting)

[mynameisfunk]

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Homework Statement

A football team consists of 20 offensive and 20 defensive players. Th players are to be paired to form roommates. They are paired at random. What is the probability that there are exactly 4 offensive/defensive pairs.

Homework Equations

The Attempt at a Solution

See attachment

motivation: -(4!)² ways to pair up the players.
-$$\frac{40!}{20!(2)^20}$$ total ways to pick 20 pairs
-(20 choose 4)² ways to pick the 4 players to be paired up
-$$\frac{32!}{16!(2)^16}$$ ways to pair up the rest of the guys

 

[lippyka]

-Probability p = \frac{(4!)2(20 choose 4)2\frac{32!}{16!(2)^16}}{\frac{40!}{20!(2)^20}} = \frac{256\frac{32!}{16!(2)^16}}{\frac{40!}{20!(2)^20}} =\frac{256\frac{32!}{16!(2)^16}}{40\cdot 39 \cdot \frac{38!}{20!(2)^20}} =\frac{128\cdot 38!}{20!\cdot 19!\cdot 18!}