2017 AMC 10A Problem 19

Alice is the most restricted person: whoever sits beside her must be Derek or Eric. Split by Alice's chair, numbering the chairs $1$ to $5$.

Alice in chair $3$ (middle). Chairs $2$ and $4$ hold Derek and Eric in some order ($2$ ways). They are not adjacent to each other, so no conflict. Bob and Carla take chairs $1$ and $5$ ($2$ ways). Total $2 \cdot 2 = 4$.

Alice in chair $2$. Chairs $1$ and $3$ hold Derek and Eric ($2$ ways); they are separated by Alice. Bob and Carla fill chairs $4$ and $5$ ($2$ ways). Total $4$. By left-right symmetry, Alice in chair $4$ also gives $4$.

Alice in chair $1$. Chair $2$ is Derek or Eric ($2$ ways). Say Derek; then Eric cannot sit in chair $3$. Chairs $3, 4, 5$ hold Bob, Carla and Eric with Eric in chair $4$ or $5$ ($2$ ways) and Bob, Carla in the remaining two chairs ($2$ ways). Total $2 \cdot 2 \cdot 2 = 8$. By symmetry, Alice in chair $5$ also gives $8$.

Adding: $4 + 4 + 4 + 8 + 8 = 28$.

The answer is $\boxed{\textbf{(C)}\ 28}$.

When one person is forbidden from two of the four others, fix that person first: her neighbors are forced to come from a two-person pool, which collapses the count. Symmetry (reversing the row) halves the casework. The Derek-Eric rule only bites when Derek and Eric can end up side by side, which happens only when Alice has a single neighbor.

• Using inclusion-exclusion on three forbidden adjacencies and losing track of the overlaps, or forgetting the Derek-Eric restriction when Alice is at an end.
• Counting the end cases as $2 \cdot 3! = 12$ each (ignoring Eric's restriction), which gives $36$.