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  4. The number of ways in which 5 dance couples (consisting of 1 man and 1 woman in each dance couple) can be formed from 5 married couples (including Naresh and Rani) such that no husband dances with his own wife but Naresh wants to dance with Rani (not his wife)

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Q. The number of ways in which 5 dance couples (consisting of 1 man and 1 woman in each dance couple) can be formed from 5 married couples (including Naresh and Rani) such that no husband dances with his own wife but Naresh wants to dance with Rani (not his wife)

Permutations and Combinations

Solution:

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Case-I: $H _1$ dance with $W _2$ and $H _2$ dance with $W _1$ then required case
$=3:\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}\right)=3-1=2$
Case-II: When $H _1$ dance with $W _2$ and $H _2$ will not dance with $W _1$.
$=4 !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)=12-4+1=9$
Total = 11