Suppose that we have a (regular) tetrahedron and that we are going to paint the 4 triangular faces of the tetrahedron using red, white, blue and yellow paint. How many different ways can you do that?

Well if you've ever done any combinatorics (or even if you haven't) you'll jump up and say "I know I know. The answer is easy, there are 4 ways to paint the first face, 4 ways to paint the second, 4 ways for the third and 4 ways for the last. Altogether that gives 44=256 ways."

Of course you'd be wrong, wrong, wrong! Why? Well, suppose I coloured my tetrahedron somehow. Now grab hold of one of the vertices and then rotate the tetrahedron through 2pi/3 about the axis through that vertex and the centre of the tetrahedron. That rotation will leave the tetrahedron shape unchanged but will move the colouring I started with into another one. But I don't have any right to call this colouring different from the first one. It's just the same.

In other words the 256 ways I counted includes some repetitions. For the real answer to the question I have to take all the rotations of the tetrahedron into account.

Well first of all what are the rotations of the tetrahedron? As mentioned in the symmetry groups of the Platonic solids this group is isomorphic to the alternating group A4, in particular it has 12 elements. Let's write them down. I just described a rotation about 2pi/3 about an axis through a vertex and the midpoint of the opposite face. Likewise we could rotate through 4pi/3. Now there are 4 vertices so altogether this gives use 8 elements of the group. That leaves 4 to go. One is easy, just the identity (which leaves everything where it is). Finally consider the line through the centre of the midpoint of an edge and the centre of the tetrahedron and out through the midpoint of the opposite edge. we get a rotation through pi about this axis. There are six edges, so three such edge pairs, giving us the final three elements of the rotation group.

Now we need to use some technology. Write G for the group of rotations of the tetrahedron and X for the 256 paintings with repetitions. When we apply a rotation from G to the paintings in X then we get all the paintings back again but in some different order. In mathematical terms we have a group acting on a set (read that writeup now!). The real answer to our question is the number of orbits for this action. Luckily we can use the fixed point formula for a finite group acting on a finite set. It says that the number of orbits is given by the formula (1/|G|) Sum (g in G) |Xg| where Xg means the elements of X that are fixed by g.

Let's compute this general formula in our special case.

• |X1|=|X| (the identity fixes everything)
• If g is a rotation through 2pi/3 or 4pi/3 about the axis through a vertex and the centre then to fix a painting it must be that the three faces that meet at the vertex are all the same colour and the other face can be any colour. Thus there are 4.4=16 such paintings.
• If g is the rotation through pi about the axis through the midpoints of opposite edges then the two faces that meet at each of the two edges must be the same colour. So again there are 4.4 ways to do this.
So the formula tells us that there are 1/12(256+8.16+3.16)=36 orbits.

There are exactly 36 ways to paint a tetrahedron with 4 colours.