# 2015 AMC 10A Problems/Problem 10

## Problem

How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$. $\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

## Solution 1

The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an $a$, we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since: $acbd$ is not allowed because of the $cb$, and $acdb$ is not allowed because of the $cd$.

We get the same problem if we start with a $d$, since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$.

If we start with a $b$, the next letter would have to be a $d$, and since we can put an $a$ next to it and then a $c$ after that, this configuration works. The same approach applies if we start with a $c$.

So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$, giving us: $$1 + 1 = \boxed{\textbf{(C)}\ 2}.$$

## Solution 2 (Casework)

Case 1: the first letter is A

Subcase 1: the second letter is C

The next letter must either be B or D, both of which do not satisfy the conditions.

Subcase 2: the second letter is D

The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other.

Case 2: the first letter is B

The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works.

Case 3: the first letter is C

The next letter is forced to be A, the third letter is D and the last letter is B. This works.

Case 4: the first letter is D

Subcase 1: the second letter is A

The third letter has to be C and the fourth has to be B but this doesn't work because B and C are next to each other.

Subcase 2: the second letter is B

The third letter cannot be A or C, so this doesn't work.

Summing the cases, there are two that work: $BDAC$ and $CADB$ $\Longrightarrow \boxed{\textbf{(C)}\ 2}$.

~JH. L

## Video Solution (CREATIVE THINKING)

~Education, the Study of Everything

~ pi_is_3.14

~ ThePuzzlr

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 