How many different necklaces can be made using exactly three red and three black beads? Disclaimer: Not a homework question it just only for learning
I know i'm supposed to use the formula: $\frac{1}{ \left |G  \right |} \sum_{g \in G} \left | Fix(g) \right |$
And using the Burnside theorem, but I have problems running. I appreciate any explanation and help.
 A: Here $G=D_6$ that is the group of the $12$ symmetries of the regular hexagon. $G$ is made of $6$ rotations and $6$ reflections.
Now we evaluate the cardinality of the fixed point set for each $g\in G$:


*

*$g$ is the identity then $|\mbox{Fix}(g)|=\binom{6}{3}=20$ (why?);

*$g$ is the $\pm60^\circ$-clockwise rotation then $|\mbox{Fix}(g)|=0$ (why?);

*$g$ is the $\pm120^\circ$-clockwise rotation then $|\mbox{Fix}(g)|=2$ (why?);

*$g$ is the $180^\circ$-clockwise rotation then $|\mbox{Fix}(g)|=0$ (why?);

*$g$ is a reflection across a line through two opposite vertices then $|\mbox{Fix}(g)|=4$ (why?);

*$g$ is a riflection across a line through the midpoints of two opposite sides then $|\mbox{Fix}(g)|=0$ (why?);
Hence, by Burnside's formula,
$$\frac{1}{ \left |G  \right |} \sum_{g \in G} \left | \mbox{Fix}(g) \right |=\frac{20+2+2+4+4+4}{12}=3.$$
In fact, we have 1 necklace with three adjacent red beads, 1 necklace with exactly two adjacent red beads, and 1 necklace where beads are alternately colored.
P.S. If reflections are not allowed then $G=R_6$ that is the group of the six rotations of the hexagon, and, 
by Burnside's formula, the number of different necklaces is
$$\frac{1}{ \left |G  \right |} \sum_{g \in G} \left | \mbox{Fix}(g) \right |=\frac{20+2+2}{6}=4.$$
A: 
Since you can't embed images in comments I am posting this as an answer.

You can also simply draw these:

(from the Wikipedia page for necklaces, created by Wikipedia user Watchduck (Tilman Piesk)).
There are 3 bracelets (dihedral symmetry) and 4 necklaces (rotational symmetry). The fourth necklace is obtained as the mirror image of the middle one in the above picture.
