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

The first thing one would want to do is see a possible value 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:

$1 + 1 = \boxed{\textbf{(C)}\ 2}.$

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 10

Problem

Solution

See Also