Question

Find the number of ways in which a mixed doubles game can be organised from amongst $8$ married couples if no husband and wife play in the same team.

• A. $840$
• B. $1202$
• C. $1174$
• D. $1204$

Answer 1

The correct option is D $1204$

CASE $1:$ No two couples in the same game
Number of ways $={}^8{C}_2\times {}^6{C}_2\times 2$
where ${}^8{C}_2$ is the number of ways of selecting $2$ men from $8$ men,
${}^6{C}_2$ is the number of ways of selecting $2$ women from $6$ women,
$2\to$ number of ways to form a team

CASE $2:$ One couple in the same game
Number of ways $={}^8{C}_1\times {}^7{C}_1\times {}^6{C}_1$
where ${}^8{C}_1$ is the number of ways of selecting one couple,
${}^7{C}_1$ is the number of ways of selecting one man from the remaining couples,
${}^6{C}_1$ is the number of ways of selecting one women from the remaining $6$ couples

CASE $3:$ Two couples in the same team
This can be done in ${}^8{C}_2$ ways

Therefore, total number of ways are :
${}^8{C}_2\times {}^6{C}_2\times 2+{}^8{C}_1\times {}^7{C}_1\times {}^6{C}_1+{}^8{C}_2=1204$