# Permutations in a conference

1. Nov 6, 2009

### Jshua Monkoe

1. The problem statement, all variables and given/known data

At a conference of 5 powers,each deligation consists of 3 members. If each delegation sits together, with the leader in the middle, in how many ways ca the members be arranged at a round table?

2. Relevant equations

No. of ways of arranging n objects around a circle=(n-1)!
P(n,r)=n!/(n-r)!

3. The attempt at a solution

I understand that 15 people go around the table
=> no. of ways to arrange=14!
But again in their different delgations,each can be arranged 2!/(2-2)!=2ways
=> There are 2(15) ways to arrange within the delegations
.'. there are 14!(30) ways
IS THIS CORRECT?

2. Nov 6, 2009

### Abha

We can not consider 15 people as a whole since we do not have complete flexibility regarding their arrangement.
Since there are 5 groups which always individually sit together, there are 4! ways in which the groups can be organized on the table.
Now, each group can internally have 2 possible arrangements. => 2^5 arrangements in all.
i.e. Total no. of permutations = 4! * 2^5