Question

Number of ways in which $3$ men and their wives can be arranged in a line such that none of the $3$ men stand in a position that is ahead of his wife, is

• A. $3!3!$
• B. $2!\cdot 3!\cdot 3!$
• C. $3!$
• D. $\frac{6!}{2!2!2!}$

Answer 1

Answer: (D)

Treat $H_1{W}_1$ as alike objects say $XX$ $H_2{W}_2$ as $YY$ and $H_3{W}_3$ as $ZZ$
Now arrangements of $XXYYZZ$ i.e. $\frac{6!}{2!2!2!}$ is this required number of ways (think!)
Alter $:{}^6{C}_2($ for $H_1{W}_1)$ and ${}^4{C}_2($ for $H_2{W}_2)\times \times \times \times \times \times$
Hence total ways $\left({}^6{C}_2\cdot 1\right)\left({}^4{C}_2\cdot 1\right)=\frac{6!}{2!2!2!}$