Assume we have $10$ men and $10$ women, how many ways are there to pair them up into $10$ pairs with one man and one woman in each pair?
I can think of two possible approaches:
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Let $10$ women choose one of the men one by one. The first woman has $10$ different choices. The second woman has $9$ different choices and so on. The last woman only has one choice. Therefore, the final answer is $\boxed{10!}$
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Let $N = 2$ ($2$ men, $2$ women). The total combinations are $[m_1w_1, m_1w_2, m_2w_1, m_2w_2] = 4$. If $N = 3$ ($3$ men, $3$ women), the total combinations are $[m_1w_1, m_1w_2, m_1w_3, m_2w_1, m_2w_2, m_2w_3, m_3w_1, m_3w_2, m_3w_3] = 9$. Since $N$ men can be paired with $N$ women, it is simply $\boxed{N^2}$
Which, if any, of these reasons are correct?
Best Answer
Your first approach is correct.
The way to get an answer to this is to look at the problem inductively. You can pair the first woman off with 10 men. The second woman could then choose from 9 men. Going on like this you would conclude that the tenth woman could choose from 1 man. Hence your answer is going to be $10 \times 9 \times \dots \times 1 = 10!$.
For the second approach, if $N=4$ ($2$ men, $2$ woman) then the first woman can pair with 2 men while the second woman can pair with 1 man. Therefore, it would be $2!$.
As noted in the comments, for $2$ men and $2$ woman there are only $2$ possibilities
$$m_1w_1, m_2w_2\tag{1}$$ $$m_2w_1, m_1w_2\tag{2}$$
which is why it is $2$!.
If $N=6$ ($3$ men, $3$ woman) then the first woman can pair with 3 men. The second woman can pair with 2 men and the third woman can pair with one man. Therefore, it would be $3!$. This can be seen by
$$m_1w_1, m_2w_2, m_3w_3\tag{1}$$ $$m_2w_1, m_1w_2, m_3w_3\tag{2}$$ $$m_3w_1, m_2w_2, m_1w_3\tag{3}$$ $$m_1w_2, m_2w_3, m_3w_1\tag{4}$$ $$m_3w_2, m_2w_1, m_1w_3\tag{5}$$ $$m_2w_3, m_1w_1, m_3w_2\tag{6}$$