1. How many different tetrahedrons can be produced by coloring each face a solid color and using n different colors? (Two tetrahedrons are the same if they can be turned and placed side by side so that corresponding sides match in color.)

2. How many cubes with n colors?

Correction for the cube version:

The coefficients I gave for C(n,3) and C(n,4) are wrong.

For three colors, I had said "If the distribution is 2,2,2, there is just a left- and a right-handed version." This is incorrect, as it applied only to the situation where each member of a matched pair is adjacent to its mate. It's also possible that one matched pair has its members opposite each other; this is symmetric and so there's no further division of this case, except that there are 3 choices for which is the pair that holds opposing sides.  Another possibility is that the members of each of the three pairs is opposite its match. This adds a total of 3+1=4 to the coefficient of C(n,3), so that it is 30 rather than 26.

For four colors, I had said "If one doublet is adjacent but the members of the other pair are opposite, the[re] is also only one version."  However, there are two choices for which of the pairs is adjacent and which is opposite.

For four colors, I also said "But if both doublets are arranged so that matching colors are adjacent, there's a left-handed version and a right-handed version, just considering the doublets, and then there are 2 ways for each of these in regard to the placement of the remaining two colors."  This took into consideration only those cases where the 2 pairs of matching colors were not arranged in one ring around the cube, but rather nestled so that one member of each pair was adjacent to both members of the other.  So to this has to be added the case where one pair of matching colors is directly opposite another pair of matching colors.

This adds two cases, each of which is multiplied by C(4,2) (even within the C(n,4) coefficient), so we're adding 12 to the coefficient of C(n,4).  I had previously reported that coefficient as 60, but as you can check, it really should have been 56--my arithmetic was wrong.  So the new total should be 68.

So the correct formula for cubes is:

n + 8*C(n,2) + 30*C(n,3) + 68*C(n,4) + 75*C(n,5) + 30*C(n,6) ways of painting the cube all together.

This matches the formula given in the Online Encyclopedia of Integer sequences, at
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A047780