If I had a cube and six colours and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had n colours instead of 6?
I *think* that we are supposed to assume two colorings are equivalent if the cube can be rotated so that one coloring coincides with the other. If that is the case, then Mr. Fantastic's answer is too high.
OK, let's say two colorings are equivalent if the cube can be rotated so that they coincide. We would like to count the non-equivalent colorings.
Let's say one of the colors is red. Place the cube on your desktop and position the red face on top. There are then 5 ways to color the bottom face. For each of these, there are 4! / 4 ways to color the side faces if we consider the 4 rotations of the cube, leaving the top and bottom in place, to be equivalent. So there are 5 * 4! / 4 = 30 ways in all.
(There is another way to approach this problem, using generating functions and the Polya Enumeration Theorem, but that's probably more advanced than you want to read about.)
How do you know to divide by 4 for the rotation? Also, since this is equivalent to 3!, can the answer be worked out by taking (n-1)! for the n side faces, or is that just for when there are 4 side faces?