Difference between revisions of "2015 AMC 10A Problems/Problem 10"

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(Solution: Added LaTeX in a few places.)
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==Solution==
 
==Solution==
  
Observe that we can't begin a rearrangement with either a or d, leaving bcd and abc, respectively.
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Observe that we can't begin a rearrangement with either <math>a</math> or <math>d</math>, leaving <math>bcd</math> and <math>abc</math>, respectively.
  
Starting with b, there is only one rearrangement: <math>bdac</math>. Similarly, there is only one rearrangement when we start with c: <math>cadb</math>.
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Starting with <math>b</math>, there is only one rearrangement: <math>bdac</math>. Similarly, there is only one rearrangement when we start with <math>c</math>: <math>cadb</math>.
  
 
Therefore, our answer must be <math>\boxed{\textbf{(C) }2}</math>.
 
Therefore, our answer must be <math>\boxed{\textbf{(C) }2}</math>.

Revision as of 17:05, 4 February 2015

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 $ba$.

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 3\qquad\textbf{(E)}\ 4$ (Error compiling LaTeX. Unknown error_msg)


Solution

Observe that we can't begin a rearrangement with either $a$ or $d$, leaving $bcd$ and $abc$, respectively.

Starting with $b$, there is only one rearrangement: $bdac$. Similarly, there is only one rearrangement when we start with $c$: $cadb$.

Therefore, our answer must be $\boxed{\textbf{(C) }2}$.