2017 AMC 10A Problem 19

If Alice sits on an end, then the person next to her cannot be Bob or Carla. This means that it must be Derek or Eric. WLOG, let the person be Eric. Then the person next to Eric has to be Bob or Carla. After that there are no more restrictions. This gives us a total of $$2 \cdot 2 \cdot 2 \cdot 2 = 16.$$ The first $2$ is for both edges. The second $2$ is for Derek or Eric. The third $2$ is for Bob or Carla. The final $2$ is just for the $2$ people that are remaining. Otherwise, let Alice be in one of the three non-end seats. Then the two people next to her have to be Derek and Eric. Bob and Carla are forced to be in the last $2$ seats. There are $3$ choices for Alice’s seat. The side on which Derek sits has $2$ options, and then there are $2$ options for where Bob and Carla go. This gives us $$3 \cdot 2 \cdot 2 = 12$$ configurations. Therefore, there are a total of $12 + 16 = 28$ total seating arrangements. Thus, C is the correct answer.