Question

The number of ways in which a mixed double tennis game can be arranged from amongst $7$ married couple if no husband & wife plays in the same game is

• A. $210$
• B. $420$
• C. $560$
• D. $840$

Answer 1

Answer: (C)

$560$

Let the 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 $B$ can 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$=10\times 3\times 2=60$.