Question

In how many ways can a game of lawn tennis mixed double be made up of seven married couples if no husband and wife play in the same set?

Answer 1

Step 1: Calculate the number of ways to select two men

It is given that there are $7$ couples, i.e., $7$ husbands and $7$ wives.

Let us suppose, two husbands $P$ and $Q$ are selected from the $7$ husbands.

Then, the ways of selecting two husbands $P$ and $Q={}^{7}C_2$

$\Rightarrow$ the ways of selecting two husbands $P$ and $Q=\frac{7!}{2!\left(7-2\right)!}$ $\mathrm{[}\because {}^{n}C_r=\frac{n!}{r!(n-r)!}\mathrm{]}$

$\Rightarrow$ the ways of selecting two husbands $P$ and $Q=\frac{7!}{2!\times 5!}$

$\Rightarrow$ the ways of selecting two husbands $P$ and $Q=\frac{7\times 6\times 5!}{2\times 1\times 5!}$

$\Rightarrow$ the ways of selecting two husbands $P$ and $Q=21$

Step 2: Calculate the number of ways to select two women

Now, we need to select two wives (say) $R$ and $S$ such that they are not the wives of $P$ and $Q$.

Then, the number of remaining wives $=7-2=5$

So, the ways of selecting two wives $R$ and $S={}^{5}C_2$

$\Rightarrow$ the ways of selecting two wives $R$ and $S=\frac{5!}{2!\left(5-2\right)!}$ $[\because {}^{n}C_r=\frac{n!}{r!(n-r)!}]$

$\Rightarrow$ the ways of selecting two wives $R$ and $S=\frac{5!}{2!\times 3!}$

$\Rightarrow$ the ways of selecting two wives $R$ and $S=\frac{5\times 4\times 3!}{2\times 1\times 3!}$

$\Rightarrow$ the ways of selecting two wives $R$ and $S=10$

Step 3: Calculate the number of ways of making mixed double

Therefore, the number of ways to select the players for a mixed double can be calculated as,

The number of ways to select a team $=$ ways of selecting two husbands $\times$ ways of selecting two wives

(Where these wives are other than those of the selected husbands)
$\Rightarrow$ The number of ways to select a team $=21\times 10$
$\Rightarrow$ The number of ways to select a team $=210$
Now, let us say that $P$ chooses $R$ as a partner (then $Q$ will automatically go to $S$) or $P$ chooses $S$ as a partner, (then $Q$ will automatically go to $R$).

Thus we have $2$ choices for the teams.

So, the required number of ways $=2\times 210$

$\Rightarrow$ the required number of ways $=420$

Hence, the required number of ways is $420$.