Counting the number of colored tetrahedra The book “Mathematics of choice or how to count without counting” by Ivan Niven has the following question:
Problem set 1, question 15:
A man has a large supply of wooden regular tetrahedra, all the same size.  (A regular tetrahedron is a solid figure bounded by four congruent equilateral triangles.)  If he paints each triangular face in one of four colors, how many different painted tetrahedra can he make, allowing all possible combinations of colors?  (Say that two blocks are different if they cannot be put into matching positions with identical colors on corresponding faces.)
I attempted to solve this problem in cases by examining the numbers of possible painted tetrahedra with only one color, then two colors, ...
I think the answer is 48 but the book answer is slightly different (36).  We disagree about the case where the tetrahedra are painted with three colors.  I think there are 24 cases but the book says 12.  Who is right?
My second question is:  Can this problem be solved more elegantly (no case analysis)?
 A: If you use three colors you have four choices for the color that is on two faces.  You then have three choices for the next color for one face and two choices left for the last face, which presumably gets your $24$ by multiplying.  However, you can choose the last two colors in either order and get the same tetrahedron as the two faces painted with single colors are not in different positions, so you must divide by $2$ getting $12$
A: Burnside's Lemma tells us that we should look at every way to rotate the tetrahedron onto itself and for each rotation, count the number of coloured tetrahedra which remain unchanged after applying the rotation (fixed points). The average number of fixed points is the number of distinct colourings.
There are 12 rotations (including the identity rotation). These can be partitioned into three classes (conjugacy classes) for which the number of fixed points is the constant on the whole class:

*

*doing nothing/the identity permutation (cycle type $[1^4]$)

*a double transposition which flips one pair of faces meeting the same edge as well as flipping the second pair of faces (cycle type $[2^2]$)

*a clockwise or counterclockwise rotation around some vertex (cycle type $[31]$)

https://de.zxc.wiki/wiki/A4_(Gruppe)
If there are $n$ colours to be used ($n = 4$ in the problem) then

*

*the identity permutation fixes $n^4$ tetrahedra (any possible colouring is fixed by doing nothing)

*a double swap fixes $n^2$ (the swapped faces have to be the same colour)

*a rotation fixes $n^2$ (the three faces around the vertex must be the same colour if they remain unchanged under this rotation)

There is

*

*one identity permutation

*three double transpositions (6 ways to select a single pair divided by 2 for the symmetry [the double pair 1,2 and 3,4 is the same as 3,4 and 1,2])

*eight vertex rotations (4 vertices × 2 directions to rotate)

$1$ permutation (the identity) has $n^4$ fixed points and the remaining $11$ permutations have $n^2$ fixed points. Burnside's lemma tells us to take the average:
$$\boxed{\frac{1}{12}(n^4 + 11n^2).}$$
For $n = 4$ we get $\boxed{36}$ distinct colourings.
