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I'm looking to count ways to make $n$ pairs if there are two groups of $n$ people and each pair must consist of one person from each group.

My initial thought is $^nP_n = n!$, as we can line the groups up, fix the positions of one line, and only permute the other line to make all possible pairings.

I'd also be interested in hearing the intuition behind why this is different from picking $n$ pairs from a single group of $2n$ people, which seems to have $$\frac{(2n)!}{n!2^n}$$ possible sets of pairs.

Ryan
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    Your $n!$ is correct. The second approach is different because it removes the restriction that "each pair must consist of one person from each group." – angryavian May 26 '22 at 15:59

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For the first part, you have already given a nice intuitive explanation.

For an intuitive explanation of the second formula, in the $(2n)!$ permutations of the items which we can pair serially, neither the order of the pairs nor the order within the pairs matter, thus the division by $n!2^n$,

I much prefer, though, to think of the ways for sequential choices being $[(2n-1)(2n-3)(2n-5)...1 = (2n-1)!!$,

true blue anil
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I like your reasoning for the case of two separate groups, and I believe that it is correct: we can simply look at each of the $n$ elements of one group in order, and for the $i^{th}$ element we will have $n - i + 1$ options when we select from the other group without replacement, giving us $\prod_{i = 1}^n n - i + 1 = \prod_{k = 1}^n k = n!$ total groupings. (this isn't really necessary, I just thought that expanding the argument out a bit would help us check why it's true)

Regarding why the case for a single group of $2n$ is different, it's because the restriction of the two groups has been removed, or equivalently, we have the choice to select the groups ourselves. If we want to use a similar argument, we can consider splitting any two permutations of our $2n$ elements into two groups of $n$ and matching from the two groups as we did before.

This gives us $(2n)!$ pairs of ordered groups, but once we have our two groups we need to note that if we rearrange the two groups of $n$ in the same way then we get the same set of pairs, so we're overcounting by a factor of the number of orders of one of our lines of $n,$ which is $n!.$

This is the only way to reorder within groups without changing the pairs, but if we have that the pairs are unordered then we can also swap any two paired people between their groups without changing the pairs. For any set of pairs there are $2^n$ such ways we can choose to swap the $n$ pairs, so we also need to divide by that. Because these are the only two ways we can change the order of our $2n$ elements without changing the pairs, we get that the total number of ways to pair $2n$ people is

$$\frac{(2n)!}{n! \cdot 2^n} = (2n-1)!!$$

where the double-factorial representation matches an intuitive method of putting everyone in a line, matching the first person with any of the other $2n - 1$ people, then matching the next person with any of the remaining $2n - 3,$ and so on, matching the $i^{th}$ person considered with any of the remaining $2n - 2i + 1$ people, giving us

$$\prod_{i = 1}^n 2n - 2i + 1 = \prod_{k = 1}^n 2k - 1 = (2n - 1)!!$$

Stephen Donovan
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  • Thanks for the answer, your explanation of splitting permutations of $2n$ into two groups made the difference clearer! – Ryan May 26 '22 at 17:08