Q:

How many ways can we paint the sides of a pyramid if two adjacent sides can't have the same color?

Given $4$ different colours, we want to paint the four sides of a pyramid in such a way as to sides that share an edge do not have the same colour. We're allowed to use the same colour more than once otherwise.
I drew a pyramid viewed from the top, to help me visualize.

The solution to this problem is $4\cdot3\cdot2\cdot2+4\cdot3\cdot1\cdot3$. However, I can't quite understand why. I was hoping anyone would provide an intuitive explanation.

A:

First label the sides by their (inter)cardinal directions. We have NE, NW, SE and SW.
NW and SE either have the same color, or different colors. This splits us off into two cases:

Different color. Pick the color for NE first. There are 4 to choose from. Then pick the color for NW, there are 3 to choose from. Now, since we said SE would be different from NW, and it has to be different from NE, there are 2 colors to choose from. Finally, there are 2 colors to choose from for SW. In total, $4\cdot 3\cdot 2\cdot 2$ possibilities.
Same color. Pick the color for NE first. There are 4 to choose from. Now pick the color for NW. There are 3, since it can't be the same as NE. Then there is only 1 available choice for SE since we said it would be the same as NW, and finally, since NW and SE have the same color, there are 3 available for SW. In total $4\cdot 3\cdot 1\cdot 3$ possibilities.

Add them together, and you have the answer.

