<s>[INST] Necklace problem with Burnside Lemma How many necklaces can be made with two red beads, two green beads, and four violet beads?(8 total)
Using Burnside lemma is complicated for me due to my lack of understanding of the lemma. I want to know the method step-by-step.
 [/INST] We may as well deploy PET here since we need the cycle index $Z(D_8)$ of
the dihedral group $D_8$ of order $16$ to apply Burnside. We compute
and average the number of assignments of colors to the eight slots fixed
by permutations from  each conjugacy class in $D_8$, taking into  account
the order of the class. This means the assignment is constant on the
cycles, so we can place exactly one color in the slots on a given  cycle,
substituing $a_d$ from the index with $R^d + G^d + V^d,$ which is PET.

Consulting the following fact sheet on necklaces and
bracelets
we get for the cycle index of the dihedral group $D_8$
$$Z(D_8) = \frac{1}{16} a_1^8
+ \frac{1}{4} a_1^2 a_2^3
+ \frac{5}{16} a_2^4
+ \frac{1}{8} a_4^2
+ \frac{1}{4} a_8.$$
We seek $$[R^2 G^2 V^4] Z(D_8; R+G+V).$$
Here we assume that the OP asks for the full symmetry i.e. dihedral,
which means that the label to use is bracelet. Working through the five
terns  in the cycle index we obtain

 $\frac{1}{16} a_1^8$
$$[R^2 G^2 V^4] \frac{1}{16} (R+G+V)^8
= \frac{1}{16} {8\choose 2,2,4}$$

 $\frac{1}{4} a_1^2 a_2^3$
$$[R^2 G^2 V^4] \frac{1}{4} (R+G+V)^2 (R^2 + G^2 + V^2)^3
\\ = [R^2 G^2 V^4] \frac{1}{4} (R^2 + G^2 + V^2) (R^2 + G^2 + V^2)^3
\\ = [R^2 G^2 V^4] \frac{1}{4} (R^2 + G^2 + V^2)^4
\\ = [R G V^2] \frac{1}{4} (R + G + V)^4
= \frac{1}{4} {4\choose 2,1,1}.$$

 $\frac{5}{16} a_2^4$
$$[R^2 G^2 V^4] \frac{5}{16} (R^2 + G^2 + V^2)^4
= \frac{5}{16} {4\choose 2,1,1}.$$

 $\frac{1}{8} a_4^2$
$$[R^2 G^2 V^4] \frac{1}{8} (R^4 + G^4 + V^4)^2
= 0.$$

 $\frac{1}{4} a_8$
$$[R^2 G^2 V^4] \frac{1}{4} (R^8 + G^8 + V^8)
= 0.$$


We thus get for our answer
$$\frac{1}{16} {8\choose 2,2,4} + \frac{9}{16} {4\choose 2,1,1}
= \bbox[5px,border:2px solid #00A000]{33.}$$
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