Painting a cube with 3 colors (each used for 2 faces). 
A cube is about to get fully painted using $3$ different colors. Each
  color is being used for $2$ faces of a cube. How many different cubes
  can be created this way?

I saw this in a fifth grade math contest and it does not appear to be an easy problem. It took me almost ten minutes to figure out the answer (which happens to be $6$) but I did it by imagination and visualization which is not nearly as satisfactory as a rigid proof.
Could somebody offer an interpretation in terms of advanced mathematics? Group theory (see Burnside's lemma) or maybe graph theory is what comes to mind...
 A: This one can be done with the Polya Enumeration Theorem, which requires the cycle index $Z(G)$ of the face permutation group $G$ of the cube. We now enumerate the permutations from this group by their cycle structure.
There is the identity, which contributes $$a_1^6.$$
There are rotations about diagonals connecting opposite vertices, which contribute
$$4 \times 2 \times a_3^2.$$
There are the rotations about an axis passing through the centers of opposite faces, which contribute $$3\times (2 a_1^2 a_4 + a_1^2 a_2^2).$$
Finally there are rotations about an axis passing through the midpoints of opposite edges, which contribute
$$6\times a_2^3.$$ 
It follows that the cycle index is
$$Z(G) = \frac{1}{24} a_1^6 + \frac{1}{3} a_3^2 +
\frac{1}{4} a_1^2 a_4 + \frac{1}{8} a_1^2 a_2^2 + \frac{1}{4} a_2^3.$$
Substituting the three colors red, green, and blue into this cycle index we get
$$Z(G)(R+G+B) =
1/24\, \left( R+G+B \right) ^{6}+1/4\, \left( R+G+B \right) ^{2} \left( {R}^{4}
+{G}^{4}+{B}^{4} \right)\\ +1/8\, \left( R+G+B \right) ^{2} \left( {R}^{2}+{G}^{2
}+{B}^{2} \right) ^{2}+1/3\, \left( {R}^{3}+{G}^{3}+{B}^{3} \right) ^{2}+1/4\,
 \left( {R}^{2}+{G}^{2}+{B}^{2} \right) ^{3}.$$
Expanding this cycle index we obtain
$${B}^{6}+{B}^{5}G+{B}^{5}R+2\,{B}^{4}{G}^{2}+2\,{B}^{4}GR+2\,{B}^{4}{R}^{2}+2\,{
B}^{3}{G}^{3}+3\,{B}^{3}{G}^{2}R+3\,{B}^{3}G{R}^{2}\\+2\,{B}^{3}{R}^{3}+2\,{B}^{2
}{G}^{4}+3\,{B}^{2}{G}^{3}R+6\,{B}^{2}{G}^{2}{R}^{2}+3\,{B}^{2}G{R}^{3}+2\,{B}^
{2}{R}^{4}\\+B{G}^{5}+2\,B{G}^{4}R+3\,B{G}^{3}{R}^{2}+3\,B{G}^{2}{R}^{3}+2\,BG{R}
^{4}+B{R}^{5}+{G}^{6}+{G}^{5}R\\+2\,{G}^{4}{R}^{2}+2\,{G}^{3}{R}^{3}+2\,{G}^{2}{R
}^{4}+G{R}^{5}+{R}^{6},$$
which finally gives $$[R^2G^2B^2] Z(R+G+B) = 6.$$
This generating function includes all distributions of three or fewer colors,e.g. the coefficient of $GR^5$ is one because there is only one way to paint the cube using one color five times and a second one for the remaining face. Note that we had rotations about face pairs, edge pairs and vertex pairs, which is a standard feature in the symmetry groups of regular polyhedra. The reader is invited to compute the cycle index of the symmetry group of the faces of a tetrahedron.
Here is another interesting calculation involving a cycle index.
