Question

In how many ways can a mixed doubles game in tennis be arranged from $5$ married couples, if no husband and wife play in the same game and two males are always opponents to two females?

Answer 1

Now let sides of the game be $A$ and $B$. Given $5$ married couples,
i.e. $5$ husbands and $5$ wives. Now $2$ husbands for two sides
$A$ and $B$ be selected out of $5={}^5{C}_2=10$ ways.

After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game). So we are to choose $2$ wives out of remaining $5-2=3$ wives i.e. ${}^3{C}_2=3$ ways.

Again two wives can interchange their sides $A$ and $B$ in $2!=2$ ways.

By the principle of multiplication. The required number of ways $=10\times 3\times 2=60$ ways.