[Math] Combinatorics : dividing people into groups

combinatorics

A group of 7 people is going to be divided in 4 groups, three of which of size 2. The order of the people within a group does not matter and groups can only be told apart in view of their sizes. How many ways are there to do so?

I know this is an easy question and my answer is 630 but I can't help but think the meaning of the phrase "groups can only be told apart in view of their sizes".

I sincerely hope it doesn't change the answer.

Best Answer

I am afraid the answer does change, because $3\;$ of the groups are indistinguishable (unlabelled).

The answer thus should be $\dfrac{630}{3!} = 105$

The reason may be clearer if we consider one such division
We would consider $AB\; |\; CD \;|\; EF\;|\;G\; $, and, say $\;CD\; |\;EF\;|AB\;|\; G$ to be identical divisions.

Using permutations, we would write the answer as $\;\dfrac{7!}{(2!2!2!1!)\times(3!)}$

The denominator's first part removes permutations within groups,
and the $3!$ removes those between groups