# Coloring of a Rhombic Dodecahedron

I need to find the number of ways to color a rhombic dodecahedron using 3 colors. I know that I need to start by working out the group of rotations, and I have seen a similar post for a regular dodecahedron. I would like to use Burnside's Lemma to show a general formula for n-given color schemes for a rhombic dodecahedron. Can anyone assist with the computation?

• I'm not sure why this was closed by a bot. The question is about a rhombic dodecahedron, the alleged duplicate is for an ordinary dodecahedron. These are different objects. – ahulpke May 1 at 21:11

We can see that the $$12$$ faces of a rhombic dodecahedron are in one-to-one correspondence with the $$12$$ edges of the inscribed cube, whose $$8$$ vertices coincide with the $$8$$ threefold vertices of the rhombic dodecahedron.

With this in mind, we clearly see that the group of rotations of the rhombic dodecahedron is the same as the group of rotations on the cube, but the action of the group is not on the six cubic faces, but the $$12$$ cubic edges. So the cycle index polynomial for the automorphisms of the rhombic dodecahedron's faces is equal to the cycle index polynomial for the automorphisms of the inscribed cube's edges.

Explicitly, for the $$12$$ rhombohedral faces:

• $$1$$ identity permutation, fixing all $$12$$ faces: $$a_1^{12}$$.
• $$6$$ rotations by $$180^\circ$$ through the centers of opposing faces, which fix the two opposing faces, and the other $$10$$ faces permute in $$5$$ pairs: $$6a_1^2 a_2^5$$.
• $$8$$ rotations by $$120^\circ$$ through opposing threefold vertices, which permute $$4$$ triplets of faces: $$8a_3^4$$.
• $$3$$ rotations by $$180^\circ$$ through opposing fourfold vertices, which permute $$6$$ pairs of faces: $$3a_2^6$$.
• $$6$$ rotations by $$90^\circ$$ through opposing fourfold vertices, which permute $$3$$ quadruplets of faces: $$6a_4^3$$.

This yields the cycle index polynomial $$Z_G(a_1, a_2, a_3, a_4) = \frac{1}{24} \left(a_1^{12} + 6 a_1^2 a_2^5 + 8 a_3^4 + 3 a_2^6 + 6 a_4^3\right).$$

I now leave it to you to finish the enumeration via Burnside's lemma/Pólya's enumeration theorem.

With the cycle index computed here's how to get colorings with exactly $$Q$$ colors (maximum of $$Q$$ is twelve):

$$\frac{Q!}{24} \left( {12\brace Q} + 6 {7\brace Q} + 8 {4\brace Q} + 3 {6\brace Q} + 6 {3\brace Q}\right)$$

We partition the cycles of a face (edge) permutation into $$Q$$ non-empty sets, one for each color. As required by Burnside the colors are constant on the cycles. The Stirling numbers of the second kind give the partition into $$Q$$ sets but these sets can be matched to the $$Q$$ colors in $$Q!$$ ways, accounting for the multiplier $$Q!$$ on all terms.

We get the finite sequence

$$1, 216, 22164, 613804, 6901425, 39713430, 131754420, 267165360, \\ 336798000, 257796000, 109771200, 19958400, 0, \ldots$$

so that in particular for three colors we obtain $$22164$$ colorings. This sequence is at OEIS A338142 where we find confirmation of these data.