Question

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)

• A. 44
• B. 11
• C. 65
• D. 205

Answer 1

Answer: (B)

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