# Application of Burnside's lemma to necklaces of three colors

I am trying to figure out the following problem using Burnside's lemma/formula:

How many different necklaces can we make using 12 equally spaced stones if we have 4 red, 5 green and 3 blue beads?

I am stuck when trying to find the fixed points under each action in D_{24}

I have so far found that if we let $$D_{24} = where x is a flip and y is a rotation of the necklace then

$$|Fix(e)| = \frac{12!}{3!4!5!}$$

$$|Fix(y^j)| = 0 \forall j$$

where $$Fix(\cdot)$$ is the set of elements in the set of necklaces that are fixed under the group action of an element in $$D_{24}$$

I am stuck trying to find $$Fix(x^j)_{j \leq 2}$$ and $$|Fix(xy^j)|_{j \leq 12}$$

any hints on how to approach this problem would be apprecaited

## 1 Answer

You've worked out the number of fixed points for every rotation. The other elements of $$D_{24}$$ are flips.

There are two kinds of flips: half are around an axis that pass through no beads, and half are around an axis that passes through two beads. The former are easy, the latter need some counting.