Question

If q colors are available, how many ways can
a. the vertices of a tetrahedron be painted?
b. the edges of a tetrahedron be painted?

Answer 1

Since a tetrahedron has only 4 faces, and you have only two colors of paint. you can just as easily count them by hand, and in fact there are just 5 distinguishable ways to paint the faces. However, you want to see how to do it using Burnside's formula.

Burnside's formula says, count the number of invariant colorings for each rotation, add up those numbers, and divide by the number of rotations. In other words, compute the average number of invariant colorings for a random rotation.

The tetrahedron has 12 rotations: you can rotate it 120oeither way about an axis passing through a vertex and the center of the opposite face, 8 of those; or 180o about an axis through two opposite edges, 3 of those; or the 0o identity rotation. Take them case by case.

A 120o rotation about an axis through a vertex vv: for the coloring to be invariant under this rotation, the three faces adjacent to vv have to be the same color; in other words the faces are partitioned into 22 "orbits", one orbit consisting of the three faces adjacent to v, the other orbit containing the remaining face. The number of invariant colorings is 2∗2=4. (If your palette contained c colors, the number of invariant colors would be c2c2.) 8∗4=32.

A 180orotation about an exis through two opposite edges, e and f: as it happens, in this case too there are 22orbits; the two faces meeting at ee belong to one orbit, so they have to be the same color; likewise the two faces meeting at ff. Again, there are 4 invariant colorings for each of these rotations. 3∗4=12.

The identity rotation: each face is a separate orbit, all 24=16 colorings are invariant, 1∗16=16.

The final answer to your question is (8∗4+3∗4+1∗16)/12
=5. For coloring the faces of a regular tetrahedron with cc colors, the number of distinguishable patterns is {11c^2+c^4}/12