On a certain block, there are five houses in a line, which are to be painted red or green. If no two houses next to each other can be red, how many ways can you paint the houses?
We break this into cases based on how many houses are red:

Case 1: No houses are red. There is only one way to do this, since all the houses must be green.

Case 2: One red house. There are 5 ways to choose the red house, and all the rest must be green.

Case 3: Two red houses. We consider where the first red house is. If it is the first house on the block, then there are 3 choices for where the second red house is (the 3rd, 4th, or 5th house), if it is the 2nd house then there are 2 choices (the 4th or 5th house), and if it is the 3rd house there is only one choice (the 5th house). Thus, there are $3+2+1=6$ options in this case.

Case 4: Three red houses. The only option is the 1st, 3rd, and 5th houses red and the other green.

Adding these together, we get $1+5+6+1=13$ ways total.