Symmetries of Tetrahedral Dice I am trying to find the number of distinguishable tetrahedral dices, where the sides are numbered 1,2,3,4. I found this webpage (http://mathworld.wolfram.com/PolyhedronColoring.html) that claims that there are 2 distinct ways to do so, but I don't see how they came to that answer. I have been trying to use Burnside's Theorem, but I have been unable to figure out how to apply it correctly. I believe that the order of the group here is $24$, but I have been having trouble computing the number of dice fixed by each group action.
 A: Start with a blank regular tetrahedron.   Label any one side with 1 and place it face down.   Label any other side 2, and rotate it away from you.  
How many ways can you label the remaining two sides with 3 and 4 to give rotationally distinct results?
It is that simplistic.
A: We can answer the questions about the number of colorings of the faces
of a tetrahedron using some  maximum number of colors or some specific
number of colors using the cycle index of the symmetries acting on the
faces of  the tetrahedron  and applying Burnside.  The cycle  index is
quite simple  here, we now show  how to compute it  by enumerating the
types of permutations.  Start with the rotations.  First  there is the
identity which contributes
$$a_1^4.$$
Next we  have rotations  by $120$ degrees  and $240$ degrees  about an
axis  passing through a  vertex and  the center  of the  opposite face
which fixes that face for a contribution of
$$4\times 2a_1 a_3.$$
Finally we  have three  $180$ degree rotations  about an  axis passing
through the midpoints of opposite edges, getting
$$3\times a_2^2.$$
Now for the  reflections, there is the first  type which exchanges the
vertices of  an edge and fixes the  faces incident on that  edge for a
contribution of
$$6\times a_1^2 a_2.$$
Lastly  there is  a type  of  reflection exchanging  the midpoints  of
opposite edges followed  by a $90$ or $270$  degree rotation about the
axis linking those two midpoints for a contribution of
$$3\times 2 a_4.$$
This last class is the most difficult and may require making a diagram
of  the  tetrahedron with  the  faces  labeled  before and  after  the
reflection  and  rotation  is  applied  and  factoring  the  resulting
permutation  by converting the  map from  table form  to a  product of
disjoint cycles, just one cycle in this case.
We thus have the cycle index
$$Z(G) = \frac{1}{24}
(a_1^4 + 8 a_1 a_3 + 3 a_2^2 + 6 a_1^2 a_2 + 6 a_4).$$
We recognize  the cycle  index $Z(S_4)$ of  the symmetric  group $S_4$
acting on four  elements. We could have noted  that given a four-cycle
(second type  of reflection)  we may choose  two elements  adjacent on
that cycle to form  a transposition (first reflection). Together these
two generate  all of $S_4,$ the  cycle index of which  can be computed
recursively  (Lovasz)  or  by  enumeration of  the  conjugacy  classes
(partitions of $n=4.$) 
Applying Burnside we get for the  number of colorings with at most $N$
colors
$$\frac{1}{24}(N^4 + 11 N^2 + 6 N^3 + 6 N)$$
which produces the sequence
$$1, 5, 15, 35, 70, 126, 210, 330, 495, 715, \ldots$$
which is OEIS A000332.
We also have for colorings using exactly $M$ colors
$$\frac{M!}{24}\left({4\brace M} + 11 {2\brace M} 
+ 6 {3\brace M} + 6 {1\brace M}\right)$$
which yields the finite sequence
$$1, 3, 3, 1, 0,\ldots $$
so there is just one coloring using four colors. This is correct since
with all  colors different we  have $4!$ possible assignments  and all
orbits have  the same size,  namely $24$, the number  of permutations,
for a  result of  $4!/24$ or one  possibility. We get  three colorings
using exactly two  colors, this corresponds to one  coloring using two
instances of each  and two colorings using three  instances of one and
one instance of the other.
