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There're 10 men and 10 women. We need to divide them into couples of same sex.

The answer is: $$\left( \frac{10!}{5!*2^5} \right)^2$$ I don't understand why.

I know that we can divide into couples where there's exactly 1 man and 1 woman in each couple in $10!$ ways (1). Now I would need to find the total possible number of couples regardless of sex $\left( \frac{20!}{10!*2^{10}} \right)$ (2). Then (2) - (1) should give the answer. But that doesn't give the same result.

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    $\begingroup$ You need to find couples of same sex. This means that the group of men and the group of women are actually independent of each other. Hence you focus on one such group, You will recognize that the number of pairs is equal to $9*7*5*3*1$. Which can be rewritten (as was done in your problem). Finally the result gets squared (equal number of combinations from men and women). $\endgroup$
    – M. Wind
    Dec 5, 2016 at 16:13
  • $\begingroup$ @M.Wind why squared and not * 2 ? Can't wrap my mind around it. $\endgroup$ Dec 11, 2023 at 0:14
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    $\begingroup$ @Vaggelis Manousakis. There are 9 * 7 * 5 * 3 combinations for men, and the same number for women. The total number of combinations is the product of the two. $\endgroup$
    – M. Wind
    Dec 11, 2023 at 4:23

4 Answers 4

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An easier way to get the number of possibilities : If we fix a man, we have $9$ choices for the first couple. If we fix a man again, we have $7$ choices, then $5$, then $3$.

In total, we have $9\cdot 7\cdot 5\cdot 3$ possibilities.

Same with the women, so finally we have $(9\cdot 7\cdot 5\cdot 3)^2$ possibilities.

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  • $\begingroup$ ok suppose we calculated that $9*7*5*3$ are the possibilities to arrange just men. Why don't we add the same for women but rather multiply? $\endgroup$
    – Yos
    Dec 5, 2016 at 16:22
  • $\begingroup$ @Yos For each combination for the men you have the same number of combinations for the women, so you have to multiply $\endgroup$
    – Peter
    Dec 5, 2016 at 16:29
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Although you didn't write the formula for the answer, it's not too difficult to guess it. For just one group of $2x$ persons, we get:

$$c(x) = \frac{2x!}{x! \cdot 2^x}$$

For the example with $10$ men and $10$ women, this would be $(c(5))^2$

And that gives you the result for all combinations as well, as you wrote $\frac{20!}{10! \cdot 2^{10}}$. But don't subtract anything from that - there is no reason to.

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$10!$ is the number of different ways to arrange the women.

Each arrangement determines the coupling (1st with 2nd, 3rd with 4th, etc).

Divide by $5!$ in order to eliminate different arrangements of each coupling.

Divide by $2^5$ is in order to eliminate different arrangements of each couple.

Raise the result to the power of $2$ in order to calculate the same for the men.

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You are right about the total possible number of couples regardless of sex. The number $\frac{20!}{10!*2^{10}}$ includes every possibility: out of the 10 couples, they can be of the same gender, or some of them can be of 1 man and 1 woman. The number $10!$ represents the case in which none of the couples are of the same sex, so you can subtract $10!$ from that to get the answer.

We first divide them into two groups of 10 persons by sex. In each of these two groups, we have $\binom{10}{2}$ ways to choose the first couple while leaving $8 \times 2$ persons behind, then we have $\binom{8}{2}$ ways to pick another couple while leaving $6 \times 2$ persons behind. We continue this process until everyone has its partner. Therefore, we have $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$, but we need to divide this by $5!$ since we ask for the number of ways of dividing them into couples. This gives the desired result for the group of 10 persons. Squaring this will give the required answer.

\begin{align} \frac{1}{5!}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} &= \frac{1}{5!} \frac{10\times9}{2}\,\frac{8\times7}{2}\,\dots\frac{2\times1}{2} \\ &= \frac{10!}{5!*2^{5}} \end{align}

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  • $\begingroup$ we need to divide by $5!$ because theoretically Anna and Mary from $\binom{2}{2}$ contribute to all other combinations? $\endgroup$
    – Yos
    Dec 5, 2016 at 16:30
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    $\begingroup$ Supoose that 5 couples $\{A,B\},\{C,D\},\{E,F\},\{G,H\},\{I,J\}$ are chosen in chronological order. However all permutations of these couples $\{C,D\},\{A,B\},\{E,F\},\{G,H\},\{I,J\}$, $\{C,D\},\{A,B\},\{G,H\},\{E,F\},\{I,J\}$ should be treated as the same outcome. $\endgroup$ Dec 5, 2016 at 16:35

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