How many ways can you color the sides of a cube using three distinct colors such that no two opposite sides are painted the same color?
I tried this: Consider any side of the cube. You can color it in $3$ ways and then the other side can be colored in $2$ ways, which gives $3\times 2 = 6$ ways for each "pair" of opposing sides. Since there are $4$ such pairs of opposing sides, we get $6 \times 3 = 18$ ways.
However, I think that this overcounts the number of ways since I don't take the symmetries of a cube into account. I know that there aren't too many possibilities, and I can probably enumerate them all, but I was trying to find some clever way to count this quantity. Maybe someone knows a way.