Question

The number of ways in which a mixed double game can be arranged from amongst $5$ married couples if no husband and wife play in the same game is

• A. $56$
• B. $60$
• C. $96$
• D. None of these

Answer 1

The correct option is B

$60$

Explanation for the correct answer:

To play a match of mixed doubles, $4$ people are required. $2$ men and $2$ women

There are $5$ married couples implying there are $5$ men and $5$ women

$2$ men can be selected from the $5$ men in ${}^5{C}_2$ ways.$...\left(i\right)$

Now we have to select the $2$ women in such a way that neither of the selected men's wives are playing the game

So we do not include the $2$ wives of the selected men in the selection for the $2$ women in the game.

Hence,$2$ women can be selected out of the remaining $3$ women in ${}^3{C}_2$$...\left(ii\right)$

Now consider that $2$ men and $2$ women are selected to play the mixed doubles match say $\left(M_1,M_2\right)$ and $\left(W_1,W_2\right)$

The teams out of these $4$ people can be $\left(M_1,W_1\right)$ and $\left(M_2,W_2\right)$ or $\left(M_1,W_2\right)$ and $\left(M_2,W_1\right)$.

That is the team can be made in $2$ ways.$...\left(iii\right)$

Hence, total number of ways$=$${}^5{C}_2\times$${}^3{C}_2\times 2$

$=\frac{5!}{2!3!}\times \frac{3!}{2!1!}\times 2$

Total number of ways$=60$

Hence, option $\left(B\right)$ is the correct answer