Question

Consider the three couples in Chintu, Pintu and Mintu and their wives Chinky, Pinky and Minky. After the dinner party, one more couple Rinku and Rinki arrives to meet them and they all decide to dance. They all want to dance in pairs where each pair consists of a man and a woman. In how many ways can such pairs be formed if nobody can have their spouse as a partner? (For example, one possible way is: Chintu & Rinky, Pinku & Chinky, Mintu & Pinky, Rinku & Minky.)

Answer 1

$\text{Since there are 4 couples, then each of them will dance with each other which is }(4!)^4. \\ \text{But, nobody can have there spouse. This will reduce each person ways to 3!. Hence} \\ (3!)^4=6^4=1296 ways$