# Coloring dodecahedron

I found some months ago that there are the Polya's enumeration theorem to compute number of colorings of dodecahedron. I got interested to find how to show by using only Burnside's lemma that there are 9099 ways to color dodecahedrom by three colors. How can I do the computation?

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You need to work out what the group of rotations of a dodecahedron is (assuming you are not considering reflections), and find the number of colourings which are preserved by each element in the group. Then you can calculate the number of orbits (i.e. the number of distinct colourings up to rotation) from Burnside's lemma.

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The following MSE link shows how to compute the cycle index of the face permutation group of the dodecahedron for use with the Polya Enumeration Theorem (PET) which generalizes Burnside.

It is shown that the cycle index is $$Z(G) = \frac{1}{60} \left( a_1^{12} + 24 a_1^2 a_5^2 + 20 a_3^4 + 15 a_2^6\right).$$ This allows to derive a formula for the colorings with at most $n$ colors and that formula is $${\frac {1}{60}}\,{n}^{12}+\frac{1}{4}\,{n}^{6}+{\frac {11}{15}}\,{n}^{4}.$$ Substitituting $n=3$ into this formula we obtain for the desired number of colorings with at most $3$ colors that indeed the count is $$9099.$$

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