In how many ways can we partition $2n$ people into $n$ teams with different names of two people each and assign to each team who plays position $A$ and who plays position $B$? (For instance, the teams could be for playing tennis doubles and the positions could signify who covers which part of the court. Note that per these instructions, if Steve and Tom were called team alpha in one arrangement and team beta in another arrangement, then these arrangements would qualify as two different arrangements.)

This is what I had, and I was told it wasn't a good answer:

If we choose each team one-by-one, we get $\binom{2n\cdot n}{n}$$\binom{(2n-2)\cdot n}{n}$...$\binom{1\cdot n}{n}$.

  • 1
    $\begingroup$ It might be $(2n)!$ $\endgroup$
    – Henry
    May 5 '15 at 8:59
  • $\begingroup$ Why? Can you please explain why you think that? $\endgroup$
    – EmaLee
    May 5 '15 at 9:01

Each possible combination of position and team name is unique and there are $2n$ of these (call them labels).

So in effect you have to allocate $2n$ people to $2n$ unique labels (or the other way round).

This can be done in $(2n)!$ different ways.


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