# Combinatorics Problem about seating around round table

I'm kind of stuck on how to do this problem.

A dinner party host has unlimited numbers of plates in a different colors. How many different arrangements of colored plates are possible at a round table which seats six people?

I have done this same problem for when there are 7 seven people at the table. I understand that this arrangement is harder because 6 is not a prime number. I just don't understand on how to approach this problem now. Any insight would be much appreciated.

To review:

To set $6$ plates of $a$ patterns in a line $= a^6$

To set $7$ plates of $a$ patterns in a circle $= \frac {a^7-a}{7} + a$ How did we get that. We set 7 plates in a line, $a^7$ Either the plates are all one color ($a$ possibilities) or there is at least one plate that is different from the rest. If there is one plate different from the rest, then we need to account for the circularity of the table.

Since 7 is prime, this is the only special consideration.

6 people at the table special cases.

Every person across the table has the same plate. (2-fold symmetery) every other person has the same plate. (3-fold symmetry) everyone has the same plate. (6-fold symmetry)

Every person across the table has the same plate. arrange 3 plates, on one side of the table, and the other side is fixed. We want to exclude all the same, as we are counting it in its own catagory. $a^2(a-1) = a^3-a$

Every other person has the same plate. $a^2 - a$

Asymmetric arrangements: $a^6 - (a^3-a) - (a^2-a) - a$

$\frac {a^6 - (a^3-a) - (a^2-a) - a}{6} + \frac {a^3-a}{3}+ \frac {a^2-a}{2} + a$