Thread: Committee Seating permutation problem

1. Committee Seating permutation problem

A committee has two presidents, two vice presidents, and two secretaries. In how many distinct ways can they sit around a circular table? Each office-holder must face across the table a person who holds a different office. This means one president cannot sit across from another president, etc. Assume that members of the same office are indistinguishable.

I began by finding the total number of arrangements, which I *think* is 120/(2!2!2!) which is a circle permutation for 6 objects divided by 2! three times because the three offices have 2 members each that can be switched around to produce the same thing. But now that I think about it I could also switch the presidents AND vp's or even all of the members.

Then, finding the number of ways we can make it so at least one office sits across from each other, there's 2 ways we can have all 3 pairs matched up, and if we have 2 pairs matched up there has to be 3 matched up. If we want only one pair matched up there are 2 ways to do it for the presidents matched up, 2 ways for the vp's and 2 ways for the secretairs. So I took 15-2-2-2-2 and got 7. However when I drew it out I have at least 8! Any thoughts?

2. Remember, arrangements around a circle are (n-1)!