Question

In how many ways, we can choose two teams of mixed double for tennis tournament from four couples such that if any couple participates, then it is the same team.

Answer 1

$CaseI$: When no couple is chosen
We can choose $2$ men in ${}^4{C}_2=6$ ways and hence two teams can be formed in $2\times 6=12$ ways.
$CaseII$: When only one couple is chosen in ${}^4{C}_1=4$ ways and the other team can be chosen in ${}^3{C}_1\times {}^2{C}_1=6$ ways. Hence two teams can be formed in $4\times 6=24$ ways.
$CaseIII$: When two couples are chosen
Then team can be chosen in ${}^4{C}_2=6$ ways.
Hence total ways $=12+24+6=42$ ways