Number of necklaces with 2 red beads and 2 blue beads

Question

I know I can easily do this problem by drawing the possible necklaces, but I am trying to understand how to do it so I can do problems with way more possible arrangements. I am trying to use the following formula, in the context of group theory.

$$\# \mathrm{Orbits} = \frac{1}{|G|}\sum_{g\in G}|\mathrm{Fix}(g)|$$

Where $G$ acts on some set $X$, and $\mathrm{Fix}(g) = \{x \in X :g\cdot x=x\}$

For this case I know there are $\frac{4!}{2!2!}$ possible initial arrangements of the beads but after this I am lost.

They are taken as the same necklace rotated – The Bosco Apr 18 '18 at 11:32 — The Bosco, Apr 18 '18 at 11:32
Probably this answer can appear useful. – user Apr 18 '18 at 11:37 — user, Apr 18 '18 at 11:37

score 0 · Answer 1 · answered Apr 18 '18 at 12:22

If turning the necklace over is not allowed then the group of necklace transitions is rotating by 0,1,2 or 3 places, which is isomorphic to the cyclic group $C_4$.

There are 6 different sequences of two red beads and 2 blue beads:

rrbb, rbrb, rbbr, brrb, brbr, bbrr

All 6 sequences are fixed by rotating 0 places. None are fixed by rotating by 1 or 3 places (only a sequence with all beads the same colour could be fixed by these transitions). 2 sequences are fixed by rotating 2 places - rbrb and brbr.

So

$\# \mathrm{Orbits} = \frac{1}{|G|}\sum_{g\in G}|\mathrm{Fix}(g)| = \frac{1}{4}(6+2) = 2$

i.e. there are just two different necklaces with 2 red and 2 blue beads.

You can see this more directly by rotating a necklace until there is a red bead at the 12 o'clock position. Then the other red bead is either at 3 o'clock, 6 o'clock or 9 o'clock. The first and third cases are equivalent (up to rotation) to the sequence rrbb; the second case is equivalent (up to rotation) to the sequence rbrb.

Number of necklaces with 2 red beads and 2 blue beads

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