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

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==Solution==
 
==Solution==
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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 <math>a</math>, we can on only place a <math>c</math> or <math>d</math> next to it.  Unfortunately, after that step, we can't do too much, since:
  
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<math>acbd</math> is not allowed because of the <math>cb</math>, and <math>acdb</math> is not allowed because of the <math>cd</math>.
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We get the same problem if we start with a <math>d</math>, since a <math>b</math> will have to end up in the middle, adjacent to an <math>a</math> or <math>c</math>, causing more excluded solutions.
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If we start with a <math>b</math>, the next letter would have to be a <math>d</math>, and since we can put an <math>a</math> next to it and then a <math>c</math> after that, this configuration works.  The same approach applies if we start with a <math>c</math>.
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So the solution must be <math>1 + 1 = \boxed{\qquad\textbf{(C)}\ 2}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:33, 7 April 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 $ba$.

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

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, adjacent to an $a$ or $c$, causing more excluded solutions.

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

See Also

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

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