2015 AMC 10A Problem 10

The forbidden neighbor pairs are $\{a,b\}$, $\{b,c\}$ and $\{c,d\}$. So the only letter allowed next to $b$ is $d$, and the only letter allowed next to $c$ is $a$.

A letter in one of the two middle positions has two neighbors, but $b$ has only one permitted neighbor, so $b$ must sit at an end with $d$ beside it. The same reasoning forces $c$ to sit at the other end with $a$ beside it. That leaves exactly two arrangements:
$$b\,d\,a\,c \qquad\text{and}\qquad c\,a\,d\,b.$$
Both satisfy the rule (the interior pair is $d, a$, which are not alphabetically adjacent).

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

With a small forbidden-adjacency problem, look for the most constrained object first. A letter with only one legal neighbor cannot be interior, so it must be at an end, and its position drags its partner along. Two such letters fix the entire arrangement up to reversal.

Brute force: there are $24$ permutations; those starting with $a$ must continue with $c$ or $d$, and a quick scan of all $24$ (or the $12$ up to reversal) confirms only $bdac$ and $cadb$ survive.

• Forbidding only $ab$ and $cd$ but allowing $bc$ (or forgetting that reversed pairs like $ba$ also count), which inflates the count to $3$ or $4$.
• Treating $a$ and $d$ as alphabetically adjacent (they are not) and concluding no arrangement exists.