[Math] For a game in which every pair has to play with every other pair, total number of games which will be played

Question

A certain game is played between pairs — one pair plays on a team against another pair. If there are $6$ players available, and every pair has to play with every other pair, what is the total number of games which must be played ?

MY APPROACH:
$6$ 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.The answer evaluates to $\frac{6!}{2!2!2!}.3=270$.

However this answer is wrong.Can you point out the mistake?

Answer 1

You are very close to being correct. The error is

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

This is wrong because you are counting each way of dividing them into 3 pairs in $6$ different ways. So you should divide by $6$ to get the correct number of ways to group them into 3 pairs.

Other than this, your calculation is correct.