# 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?

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 $$9099.$$