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.

Seeing as the going claim is that there are only 36 ways to paint a tetrahedron using four unique colors, I'll endeavor to count them. Below I have 'drawn' a tetrahedron for visualization purposes:

                         ,7`   |, `'VA,
                       ,7`     `\    `'VA,
                     ,7`        |,      `'VA,
                   ,7`          `\         `'VA,
                 ,7`             |,           `'VA,
               ,7`               `\       ,..ooOOTK`
             ,7`                  |,.ooOOT''`    AV
           ,7`            ,..ooOOT`\`           /7
         ,7`      ,..ooOOT''`      |,          AV
        ,T,..ooOOT''`              `\         /7
       `'TTs.,                      |,       AV
            `'TTs.,                 `\      /7
                 `'TTs.,             |,    AV
                      `'TTs.,        `\   /7
                           `'TTs.,    |, AV
There are two ways to paint a T (tetrahedron) with a unique color for each side; one is the mirror of the other.

There are 24 ways to paint a T with three unique colors. Four possibilities for the two that are the same, three for the first unique, and two for the second unique = 4*3*2 or 24. It is not possible to create duplicates, since duplicates are taken care of by successively reducing the number of possible colors for the next consecutive side.

There are 12 ways to paint a T where three sides are the same: 4 possibilities for the three same sides, three remaining possibilities for the remaining side for a total of 4*3=12. Again, duplicates do not exist.

There are four ways to paint a T where all the sides are the same (if it is imagined that four is the total amount of colors available).

Now, to total the results: 2+24+12+4=42. Hmm. I don't count 36. Theory and reality not quite in sync, or are my figures wrong? Just a side note, if mirror images aren't allowed, this doesn't quite cut the numbers in half, but close. They would become 1+12+12+4=29.

Update: 15:51 Tue Aug 22 2000

In reply to mblase, umm, I still have to disagree, and even add some more to my list. Remember, a mirror image is one such that no rotation can achieve it while still within the same dimension (i.e. a left hand cannot be rotated into a right hand while still in the third dimension). So my original count for three unique colors is still accurate. For two unique colors, I concede the point. So that's another six onto the original, or 48, or another six onto the secondary, or 35. I have done the work on paper and I believe it stands as I have stated it.

Update again.

Now I see the light. It took quite a bit of imagination (something I have trouble with . . .). The count is 36. So much for counting accuracy . . . The operative case here is a tetrahedron viewed on an edge. Two unique colors are on the faces visible, a third unique color is on the faces that aren't. The two visible faces can be rotated either way 180 degrees to produce a mirror image (so much for my mirror "logic" above), so there are only 12 instead of 24, and the count is 36 (or 35 if mirrors aren't allowed . . .;).
Final note: I don't intend to "fix" the mistakes above . . rather I wish this to represent a process of thought that may (or equally well may not) be useful to someone wishing to understand. My mistakes above are legitimate; I really did think I was right for quite some time (an hour at least . . .).

There are exactly two ways to paint a tetrahedron with exactly four colors. Because of how a tetrahedron rotates, one of these paintings will be an exact mirror of the other.

With three unique colors, you must have two sides the same color. (And because it's a tetrahedron, these two sides will always be adjacent.) You have four unique colors for these two sides, three for the third side, and two for the fourth. But, half of these tetrahedra will be rotations of the other half (the third side red and the fourth blue is identical to the third side blue and the fourth red, if you view it on an edge and spin it 180 degrees), so you need to divide by two: 4*3*2/2 = 12.

Where three sides are the same, you can have four unique colors for those three sides, plus three for the fourth: 4*3 = 12.

You can also have two sides be one color (out of four), and two be a different color (out of three). There are 4*3/2 (rotations again) = 6 possibilities here.

Finally, there are four colors you can use to paint all sides the same. The final total is 2+12+12+6+4 = 36 possible colorings.

Q: How many different ways can you paint a tetrahedron using no more than 4 colors?
A: 36

Unfold a tetrahedron so you get four triangles in the fashion of a first iteration Sierpinski triangle.

            /  \              
           /    \             1234  1234  1234  1234
          /      \            BBBB  BBBG  BBBR  BBBY
         /    1   \           BBGG  BBGR  BBGY  BBRR
        /          \          BBRY  BBYY  BGGG  BGGR
       /____________\         BGGY  BGRR  BGRY  BGYR
      /\            /\        BGYY  BRRR  BRRY  BRYY
     /  \          /  \       BYYY  GGGG  GGGR  GGGY
    /    \    2   /    \      GGRR  GGRY  GGYY  GRRR
   /      \      /      \     GRRY  GRYY  GYYY  RRRR
  /    3   \    /    4   \    RRRY  RRYY  RYYY  YYYY
 /          \  /          \   

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