## probability problem (counting)

1. The problem statement, all variables and given/known data

A football team consists of 20 offensive and 20 defensive players. The 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.

2. Relevant equations

3. 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 (that is a 2^20 where my graphic messed up)

-(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 (that is a 2^16 where my graphic messed up)

What confuses me is the pairing up of the players of the offensive/defensive pairs. Should it be 4! or 4!2 ? Otherwise I think the solution is good?
Attached Thumbnails

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 Quote by mynameisfunk $$\frac{32!}{16!(2)^{16}}$$ ways to pair up the rest of the guys
This is not what you want to use I think, it is the way to pair up 32 guys but it includes for example:

10 def/def pairs and 6 off/off pairs (which would mean the other 4 pairs would also be off/off)
and that would not be a case that you want to count as one that gives you the desired outcome.

 Your saying that $$\frac{32!}{16!(2)^16}$$ is the total number of ways to pair these guys up but including the rest of the off/def pairings? Also, how did you fix the exponent '16'?

## probability problem (counting)

I am saying it is including ways to make 16 pairs such that there cannot be any def/off pairs at all (since when there are 10 def/def pairs all the def players are used).

 Well isn't that what I want? I figured i would multiply the two terms A, and B where A={number of possible ways to get 4 off/def pairs} and B={number of ways the rest of the guys can get paired together} since the 2 are independent of each other. and divide by C where C={total number of ways all 40 players can be paired up}. Maybe I am misunderstanding you. PS thanks for helping it seems like noone likes my probability questions very much
 What I meant was something like this: you need to multiply A={number of possible ways to get 4 off/def pairs} with B={number of ways to get 16 pairs of either def/def or off/off}
 In my numerator could I just subtract 16!=the number of ways that the off def guys can pair with each other?

 Quote by mynameisfunk Also, how did you fix the exponent '16'?
oh missed this, use x^{16}, if you use x^16 only the 1 will be raised

 I would just go for A={number of possible ways to get 4 off/def pairs} B={number of ways to get 16 pairs of either def/def or off/off} C={total number of ways all 40 players can be paired up} and then calculate the chance as AB/C There are probably other ways, but you really need to think hard to make sure that there is no mistake somewhere when using lots of different factors... it's always easy to count something double or oversee something similar with these probability questions.
 How is this Attached Thumbnails
 Seems quite reasonable to me

 Tags bayes law, counting, data analysis, probability, statistics