# A football team consists of 20 defensive and 20 offensive players

1. Sep 27, 2010

### mynameisfunk

Hey guys, test review problem, I have the solution but having a very hard time wrapping my head around it.
Q: A football team consists of 20 defensive and 20 offensive players. The players are
to be paired to form roommates and the pairing is done at random.
(a) What is the probability that there are no offensive-defensive pairs?
(b) And that there are exactly 4 offensive-defensive pairs?

A: (a) I know that there are $$\frac{40!}{2!^20)}$$ total ways to pick pairs. Then we have $$\frac{20!}{2!^10}$$ total ways that the offensive (or defensive) players can be lined up. Soo... without looking at my solution, my best attempt is $$\frac{20!^2}{40!}$$ Having squared the numerator and simplifying. Reasoning was that the multiplication rule held for the total ways offensives can pair and total way defensives can pair over the total amount of possible pairs.

B: (b) Again, the total possible pairings are $$\frac{40!}{2!^(20)}$$. Now, we have 20C4 ways each group can pick their 4 people to be paired. And $$\frac{8!}{2!^4}$$ ways to pair these guys up. Now for the rest of one particular group we have $$\frac{16!}{2!^8}$$ ways for them to pair up, so since there are 2 of these, we square this. My final answer: 20C4$$\times$$$$\frac{16!^22!^2}{40!}$$

Last edited: Sep 27, 2010
2. Oct 2, 2010

### lanedance

Re: probability

are you sure about a) ?

have you considered the both the ordering of different pairs & the order within a pair does not change the outcome... though your tex is a little difficult to read

to explain, consider 1234, selecting numbers at random, the following pairings are all equivalent
12 34, 34 12
21 34, 43 12
12 43, 34 21
21 43, 43 21