3

So we have $N$ men and $N$ women. We are creating groups of pairs. It is not necessary to use every man and woman. How many groups can we make ?

So if we number them from $1$ to $N$ - let $W_{1}$ be woman number one and $M_{1}$ be man number 1 - we have $N$ groups like for example $[(W_{1}, M_{k})]$ where $k = {1,2...n}$.So for every woman and man we have $n * n$ groups containing a pair.

Then $[(W_{1}, M_{k}), (W_{2}, M_{l})]$. Where $k = 1,2,3,4..n$ and $l = 1,2,3..n-1$ So there is ${n \choose 2} * n * (n-1)$ possible groups of two pairs. etc. Is my reasoning correct? Is the the answer $ {n \choose k} * n * n!$ ? where $k= 1,2,3...n$

lvi
  • 289

3 Answers3

2

Your reasoning is fine. To make $k$ pairs, you can choose the $k$ women, put them in order, choose the $k$ men, and choose the order of the men, so ${n\choose k}^2k!$

Ross Millikan
  • 374,822
2

It seems like a pair has to consist of one man and one woman.

Say we want to count the number of different groups having $k$ pairs. Then we select $k$ men and $k$ women and multiply the number of ways to do each which is just $\binom{N}{k}^2$.

Assuming that the order of the pairs within the groups doesn't matter so the above expression for the number of groups with $k$ pairs is right. Otherwise you need to multiply by $k!$ per Ross's answer above.

Then just sum over all numbers of pairs per group and you get:

$$\sum_{k=0}^{N}\binom{N}{k}^2 = \binom{2N}{N}$$

A proof of this sum can be found here: Inductive proof that ${2n\choose n}=\sum{n\choose i}^2.$

As the total number of groups.

Note: If order does matter within each group then your expression for the total group count is: $\sum_{k=0}^{N}\binom{N}{k}^2k!$. Which according to Wolfram Alpha does not have a nice expression.

Community
  • 1
Patrick
  • 2,106
1

I assume that each pair must consist of a man and a woman since all example pairs are of this sort. Let there be $N$ men and $N$ women. Suppose we are making a pairing of just $M \leq N$ women

First, $W_1$ has a pairing among the $M$ men. Then $W_2$ has a pairing among the remaining $M-1$ men. Then $W_3$ has a pairing among the remaining $M-2$ men. ... Finally $W_M$ is paired with the $M-(N-1)=M-N+1$ remaining man.

Since there are $M$ initial options, then $M-1$ remaining (independent) options, then $M-2$ remaining independent options, ..., then $M-N+1$ remaining options, we compute $M \cdot (M-1) \cdot (M-2) \cdots (M-N+1) = \frac{M!}{(M-N)!}$ pairings.

Since we have ordered the women, each such assignment of men is unique. Additionally, every assignment of men is produced by this process. (Suppose not, then $W_1$ was assigned to some man and $W_2$ was assigned to some remaining man and ... and $W_M$ was assigned to one of the remaining men, so the supposed assignment was actually produced by the process.) Therefore, this counts the total number of assignments for a fixed M.

The total number of pairings is then $\sum_{M=1}^N \frac{M!}{(M-N)!} = N! \sum_{M=1}^N \frac{M!}{(M-N)!N!} = N! \sum_{M=1}^N \binom{M}{N} = N!\left(-1+ \sum_{M=0}^N \binom{M}{N} \right) = N!(2^N-1)$.

Eric Towers
  • 67,037