Question

The number of ways in which a mixed doubles game can be arranged from amongst $n$ couples such that no husband and wife play in the same game is:

• A. ${}^n{P}_4$
• B. ${}^n{C}_4$
• C. $\frac{1}{2}{.}^n{P}_4$
• D. $\frac{1}{2}{.}^n{C}_4$

Answer 1

The correct option is C $\frac{1}{2}{.}^n{P}_4$

A mixed doubles game involves two males and two females.

Two males can be chosen from $n$ males in ${}^n{C}_2$ ways.

Having chosen two males, now $2$ females are to be chosen from $(n-2)$ females leaving the wives of the two already chosen males.

This can be done in ${}^{n-2}{C}_2$ways.

Hence, the required number of ways

${=}^n{C_2.}^{n-2}{C}_2.2$ $[$for every choice of $2$males and $2$females, they can be paired in $2$ways$]$

$=\frac{n(n-1)}{2}.\frac{(n-2)(n-3)}{2}.2$

$=\frac{n(n-1)(n-2)(n-3)}{2}=\frac{1}{2}{.}^n{P}_4$