Coloring Cube

The hard way

Since rotation doesn't matter, we can fix the color of the top face as, say, red. Pick another color, say, blue. There are two possibilities:

Case 1 : Blue is adjacent to red.

Since we can rotate the cube, we can assume blue is the front face. Now the cube's orientation is fixed; there are $4! = 24$ ways to color the remaining four faces with four colors.

Case 2 : Blue is opposite of red.

Pick another color, say, green. This must be adjacent to red (there's no other available face). Using like in Case 1, we can fix the orientation by letting green to be the front face; there are $3! = 6$ ways to do the rest.

In total, that's $24 + 6 = \boxed{30}$ ways.

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The easy way

Any coloring can give rise to $6 \cdot 4 = 24$ orientations: 6 ways to fix the top face (any side can be used), and 4 ways to fix the front face (any side other than the top and the bottom faces can be used). There are $6! = 720$ ways to color the cube without taking rotation into account. We can group them into sets of 24, where in a set, all cubes can become the others by rotation. There are $\frac{720}{24} = \boxed{30}$ sets, and this is also the number of distinct colorings (that cannot be obtained through rotation).

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The overkill way

This is a way to make the "easy" way more rigorous. Let $X$ be the set of all colorings of the cube without rotation (there are $6! = 720$ such ways). Let $G$ be the group of rotations on a cube (it has $6 \cdot 4 = 24$ elements as above). Now, the number of distinct colorings on the cube is simply the number of orbits of $G$ acting on $X$ , or $|X / G|$ . This is given by Burnside's lemma :

$|X / G| = \frac{1}{|G|} \sum_{g \in G} |X^g|$

where $X^g$ is the set of elements that are fixed by $g$ ; that is, the set of colorings that don't change upon rotation by $g$ .

If there are identical colors, $X^g$ might be nonempty even for some rotation $g$ . However, because all faces have distinct colors, rotating the cube will always give a different coloring, so there is no fixed points; that is, $X^g = \emptyset$ for all non-identity $g$ . When $g$ is the identity rotation (aka no rotation), $X^g$ is simply $X$ (every element is fixed under the identity transformation). Thus we obtain:

$\begin{aligned} |X / G| &= \frac{1}{|G|} \sum_{g \in G} |X^g| \\ &= \frac{1}{|G|} \cdot |X| \\ &= \frac{1}{24} \cdot 720 \\ &= \boxed{30} \end{aligned}$

What if rotation and reflection are both count as identical? If we're using Burnside's lemma, what's the number of reflectional symmetry for the cube?