Problem:
A domino has a left end and a right end, each of a certain color. Alice has four dominos, colored red-red, red-blue, blue-red, and blue-blue. Find the number of ways to arrange the dominos in a row end-to-end such that adjacent ends have the same color. The dominos cannot be rotated.
Solution:
Without loss of generality assume that the left end of the first domino is red. Then, we have two cases:
If the first domino is red-red, this forces the second domino to be red-blue. The third domino cannot be blue-red, since the fourth domino would then be forced to be blue-blue, which is impossible. However, works.
If the first domino is red-blue, then the second domino cannot be blue-red, since otherwise there is nowhere for the blue-blue domino to go. Therefore, the second domino is blue-blue, which forces the third to be blue-red, and forces the fourth to be red-red. This yields one possibility.
Therefore, if the first color is red, there are possibilities. We multiply by to yield total possibilities.