I study maths as a hobby. I am considering this question on probability.
In how many ways can a group of 9 people attending a conference be split into 3 sets of 3 people for discussion? Later the same people form new sets of 3. In how many ways can this be done if no two people are to remain together?
I have solved the first part. The total ways of selecting people is:
$\binom {9}{3}\ . \binom {6}{3} .\binom {3}{3} = 84.20.1 = 1680$
and since the order of the groups doesn't matter we divide by 3!
$\frac{1680}{3!} = 280$
But I am stuck on the second part. I thought maybe I should first consider the number of selections in which 2 people do remain together and subtract that from 280. This would mean I would treat 2 people as one entity and make selections of 2. So I would get:
$\binom {9}{2}.\binom {7}{2}.\binom {5}{2}$ or perhaps
$\binom {9}{2}.\binom {6}{2}.\binom {3}{2}$ or perhaps
But this is getting me nowhere.
The answer is 36.
Best Answer
Each person in each original group must be placed in a separate new group.
Take one of the groups. Put each person in that group in separate groups. It does not matter how since the groups are not labeled. Say we call the new groups A, B, and C, where A is the group containing the oldest person, B is the group containing the next oldest person, and C is the group containing the youngest person from the group we split apart.
Take another of the original groups. Place one person in group A, one person in group B, and one person in group C. We can do this is in $3!$ ways.
Take the last of the original groups. Again, place one person in group A, one person in group B, and one person in group C. We can do this in $3!$ ways.
Hence, there are $3!3! = 36$ ways to rearrange the nine people who were originally placed in three groups of three people so that no two people remain together.