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How many ways are there to pair off 10 women at a dance with 10 out of 20 men?

Soln: P(20,10) X 10! (For the positions each woman could take)

My first reaponse was to use a technique with $\binom{20}{10}$ in it and then somehow get the arrangements of women. Is there a way to do it in such a manner and is my solution even correct?

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The way to get an answer to this is to look at the problem inductively. You can pair the first woman off with 20 men. The second woman could then choose from 19 men. Going on like this you would conclude that the tenth woman could choose from 11 men. Hence your answer is going to be $20 \times 19 \times \dots \times 11 = 20! / 10! = \binom{20}{10} \times 10!$.

The other way to look at this (which accounts for the last form of the answer) is to say that you could pick 10 out of 20 men to dance, which can be done in $\binom{20}{10}$ ways, and then pair off those ten men with the ten women, which can be done in $10!$ ways.

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Because it is "pair off", you should choose 1 women and 1 men from those group at the same time, so it can be (20C1)*(10C1), your answer I think just choose 10 men from the men's group but nothing with women. I think that can help

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