Counting distinguishable ways of painting a cube with 4 different colors (each used at least once) You have many identical cube-shaped wooden blocks. You have four colors of paint to use, and you paint each face of each block a solid color so that each block has at least one face painted with each of the four colors. Find the number of distinguishable ways you could paint the blocks. (Two blocks are distinguishable if you cannot rotate one block so that it looks identical to the other block.)
Having trouble solving this problem with the added constraint of "at least one face painted with each of four colors"  - Thanks in advance
 A: There is  an algorithmic approach to  this which I include  for future
reference and which consists in using Burnside and Stirling numbers of
the second  kind. For  Burnside we  need the cycle  index of  the face
permutation  group   of  the  cube.   We   enumerate  the  constituent
permutations in turn. First there  is the  identity for a contribution
of $$a_1^6.$$
Rotating about  one of the four  diagonals by $120$ degrees  and $240$
degrees we get
$$4\times 2a_3^2.$$
Rotating about an axis passing  through opposite faces by $90$ degrees
and by $270$ degrees we get
$$3\times 2 a_1^2 a_4$$
and by $180$ degrees
$$3\times a_1^2 a_2^2.$$
Finally rotating about an exis passing through opposite edges yields
$$6\times a_2^3.$$
We thus get the cycle index
$$Z(G) = \frac{1}{24}
(a_1^6 + 8 a_3^2 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 6 a_2^3).$$
As a sanity check we use this  to compute the number of colorings with
at most $N$ colors and obtain
$$\frac{1}{24}(N^6 + 8 N^2 + 12 N^3 + 3 N^4).$$
This gives the sequence
$$1, 10, 57, 240, 800, 2226, 5390, 11712, 23355, 43450, \ldots$$
which is  OEIS A047780  which looks  to be
correct.  Now if we are coloring  with $M$ colors where all $M$ colors
have to be present we must partition  the cycles of the entries of the
cycle index into a set partition of $M$ non-empty sets. We thus obtain
$$\frac{M!}{24}
\left({6\brace M} + 8 {2\brace M} + 12 {3\brace M}
+ 3{4\brace M}\right).$$
This  yields the  finite  sequence  (finite because  the  cube can  be
painted with at most six different colors):
$$1, 8, 30, 68, 75, 30, 0, \ldots$$
In particular the value for four colors  is $68.$ We also get $6!/24 =
30$ for six colors because all orbits have the same size.
A: Here's my crack at it: Starting with a stationary cube, there are $\frac{4 \cdot 3 \cdot 6!}{2 \cdot 2}$ ways of painting the cube where there are 2 colors with 2 faces, and $\frac{4 \cdot 6!}{3}$ ways of painting the cube when there is 1 color with 3 faces which gives a total of 3120 ways of painting the cube, but this over counts all the orientations of a cube, so the final answer is $\frac{3120}{4 \cdot 6} = 130$ 
A: Call the four colors $a$, $b$, $c$, $d$.
There are two partitions of $6$ into four parts, namely (1): $(3,1,1,1)$, and (2): $(2,2,1,1)$.
In case (1) we can choose the color appearing three times in $4$ ways. This color can either (1.1) appear on three faces sharing a vertex of the cube, or (1.2) on three faces forming a $\sqcup$-shape. In case (1.1) we can place the three other colors in $2$ ways ("clockwise" or "counterclockwise"); in  case (1.2) we can choose which of the three other colors is opposite the floor of the $\sqcup$. This amounts to $4\cdot(2+3)=20$ different colorings.
In case (2) the two colors appearing only once can be chosen in ${4\choose2}=6$ ways. Assume that colors $a$ and $b$ are chosen. The $a$-face $F_a$ and the $b$-face $F_b$ can be either (2.1) opposite or (2.2) adjacent to each other. In case (2.1) we can chose the two $c$-faces either adjacent or opposite to each other. In case (2.2) the two $c$-faces can be (2.2.1) opposite to each other, (2.2.2) opposite to $F_a$ and to $F_b$, or (2.2.3) opposite to one of $F_a$ or $F_b$ and on one of the faces adjacent to both $F_a$ and $F_b$. In all this amounts to $6\cdot(2+1+1+2\cdot2_*)=48$ different colorings. (The factor $2_*$ distinguishes mirror-equivalent colorings.)
Altogether we have found $68$ different admissible colorings of the cube.
