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 mixed doubles game involves two males and two females. Two males can be chosen from n males in ⁿC₂ 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}C₂ ways.

Hence, the required number of ways

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

$\begin{array}{l}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}\cdot \frac{(\mathrm{n}-2)(\mathrm{n}-3)}{2}\cdot 2\\ =\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{2}=\frac{1}{2}{\cdot }^{\mathrm{n}}{\mathrm{P}}_4\end{array}$