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mynameisfunk

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Hey guys, test review problem, I have the solution but having a very hard time wrapping my head around it.

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?

**Q:**A football team consists of 20 defensive and 20 offensive players. The players areto 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 [tex]\frac{40!}{2!^20)}[/tex] total ways to pick pairs. Then we have [tex]\frac{20!}{2!^10}[/tex] total ways that the offensive (or defensive) players can be lined up. Soo... without looking at my solution, my best attempt is [tex]\frac{20!^2}{40!}[/tex] 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 [tex]\frac{40!}{2!^(20)}[/tex]. Now, we have_{20}C_{4}ways each group can pick their 4 people to be paired. And [tex]\frac{8!}{2!^4}[/tex] ways to pair these guys up. Now for the rest of one particular group we have [tex]\frac{16!}{2!^8}[/tex] ways for them to pair up, so since there are 2 of these, we square this. My final answer:_{20}C_{4}[tex]\times[/tex][tex]\frac{16!^22!^2}{40!}[/tex]
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