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 4^{4}=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 A_{4}, 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) X^{g}
where X^{g} means the elements of X that are fixed
by g.
Let's compute this general formula in our special case.

X^{1}=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.