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This is a very basic counting problem, however I couldn't recall my memory to understand the answer correctly.
From "A First Course in Probability by Sheldon Ross",

Example A football team consists of 20 offensive and 20 defensive players. The players are paired in groups of 2 for the purpose of determining roommates....
There are $\dfrac{40!}{(2!)^{20}}$ ways of dividing 40 players into 20 "ordered pairs" of two each.

I tried a smaller set, says:

  • Offensive = ${a_1, a_2}$
  • Defensive = ${b_1, b_2}$

so all ordered pairs are : $$a_1a_2, a_2a_1, b_1b_2, b_2b_1$$ $$a_1b_1, b_1a_1, a_1b_2, b_2a_1$$ $$a_2b_1, b_1a_2, a_2b_2, b_2a_2$$ So there are 12 of pairs. Now if I use the formula was given above, I got: $$\dfrac{4!}{(2!)^2} = \dfrac{4.3.2}{4} = 6$$ which is clearly incorrect.

So my question is, should the formula given from the Example be $\dfrac{40!}{{2!}^{19}}$?

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You're confusing a single pair with a way of ordering all of the players into pairs. (Each way of ordering will consist of more than just a single pair: it will consist of enough pairs to exhaust all the team members.) Also that example seems confusing to me: $40!/2!^{20}$ counts the number of ways to split 40 players into an ordered sequence of subsets each consisting of two unordered people, whereas $40!/20!$ counts the number of ways of splitting 40 players into a set of 20 ordered pairs. – anon Sep 6 '11 at 8:24
@anon: Thanks a lot. Dunno why I don't like the wording in probability problems at all. – Chan Sep 6 '11 at 8:43
up vote 1 down vote accepted

I think you are miss understanding the thing, since it is said:

"There are $\frac{40!}{(2!)^{20}}$ ways of dividing 40 players into 20 "ordered pairs" of two each."

which it means that the ordening of the pairs in the set of pairs of a particular combination of pairs matters that pairs are ordered has not sense since the factorial would not have to be in that case with $2$), i.e. in your case would be the possibilities $$(a_1,a_2),(b_1,b_2)$$ $$(b_1,b_2),(a_1,a_2)$$ $$(a_1,b_2),(b_1,a_2)$$ $$(b_1,a_2),(a_1,b_2)$$ $$(a_1,b_1),(a_2,b_2)$$ $$(a_2,b_2),(a_1,b_1)$$ If the ordening of the pairs in the set of pairs mattered, but not the ordenind of the particular pairs, then the formula would be $$\frac{40!}{20!}$$ where the denominator represents the way of ordening the $20$ pairs.

I hope this helps.

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Thanks a lot. It does help ;). – Chan Sep 6 '11 at 8:44

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