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Question

# There are three men and seven women taking a dance class. Number of different ways in which each man is paired with a woman partner, and the four remaining women are paired into two pairs each of two is

A

$105$

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B

$315$

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C

$630$

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D

$450$

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Solution

## The correct option is C $630$Evaluate the number of ways using CombinationsGiven, $3$ women can be selected in ${}^{7}\mathrm{C}_{3}$ ways and can be paired with $3$ men in $3!$ ways.Remaining $4$ women can be grouped into two couples as $\frac{4!}{2!×2!×2!}=3$ Therefore, total number of ways$={}^{7}\mathrm{C}_{3}×3!×3\phantom{\rule{0ex}{0ex}}=\frac{7!}{\left(7-3\right)!×3!}×3!×3\phantom{\rule{0ex}{0ex}}=\frac{7!}{4!}×3\phantom{\rule{0ex}{0ex}}=5×6×7×3\phantom{\rule{0ex}{0ex}}=630$ Total number of ways$=630$Hence option $\left(C\right)$, $630$ is the correct answer.

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