2015 AMC 10A Problems/Problem 10

Revision as of 01:35, 7 April 2015 by Mattx1732 (talk | contribs) (Solution)

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

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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

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