Question

For a game in which a every pair play with every other pair, $6$ men are available.

find the number of games which can be played.

Answer 1

Solve : - $6$ players can be divided into $3$ pairs in $\frac{6!}{2!2!2!}$ ways.

Moreover $3$ pairs may play in $3$ ways with each other.

$=\frac{6!}{2!2!2!}\times 3$

$=270$.