2015 AMC 10A Problem 10

Start with the letters that have the fewest choices.

1. List the pairs that may not sit side by side: $a$ and $b$, $b$ and $c$, $c$ and $d$ (in either order).
2. So each letter's allowed neighbors are:

3. A letter in one of the two middle spots has two neighbors. Since $b$ has only one allowed neighbor, $b$ cannot be in the middle. It must be at an end, with $d$ right beside it.
4. For the same reason $c$ must be at an end, with $a$ right beside it.
5. So $b$ and $c$ take the two ends, $d$ sits next to $b$, and $a$ sits next to $c$. That leaves just
$$b\,d\,a\,c \qquad\text{and}\qquad c\,a\,d\,b.$$
6. Check the middle pair in each: $d$ and $a$ are not next to each other in the alphabet. Both arrangements work.

The answer is $\boxed{\textbf{(C)}\ 2}$.

In a "who can sit next to whom" problem, look first at the most limited item. A letter with only one allowed neighbor has to be at an end of the row, and it drags its one partner with it. Here two such letters decide the whole row.

There are only $4 \times 3 \times 2 \times 1 = 24$ orders, so you can also list them. It helps to group them by first letter. For example, starting with $a$, the next letter must be $c$ or $d$; $a\,c$ then needs a neighbor of $c$ other than $a$, which does not exist, and $a\,d$ then needs $b$ and then $c$, but $b\,c$ is not allowed. So nothing starts with $a$. Doing the same for the other first letters leaves only $bdac$ and $cadb$.