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Difference between revisions of "2015 AMC 10A Problems/Problem 10"
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How many rearrangements of <math>abcd</math> are there in which no two adjacent letters are also adjacent letters in the alphabet?  For example, no such rearrangements could include either <math>ab</math> or <math>ba</math>.
 
How many rearrangements of <math>abcd</math> are there in which no two adjacent letters are also adjacent letters in the alphabet?  For example, no such rearrangements could include either <math>ab</math> or <math>ba</math>.
  
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 3\qquad\textbf{(E)}\ 4</math>
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math>
  
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==Solution 1==
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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 <math>a</math>, we can 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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==Solution==
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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, causing it to be adjacent to an <math>a</math> or <math>c</math>.
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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 the two solutions that were allowed, one starting from a <math>b</math> and the other with a <math>c</math>, giving us:
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<cmath>1 + 1 = \boxed{\textbf{(C)}\ 2}.</cmath>
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==Solution 2 (Casework)==
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'''Case 1: the first letter is A'''
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''Subcase 1: the second letter is C''
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The next letter must either be B or D, both of which do not satisfy the conditions.
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''Subcase 2: the second letter is D''
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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.
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'''Case 2: the first letter is B'''
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The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works.
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'''Case 3: the first letter is C'''
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The next letter is forced to be A, the third letter is D and the last letter is B. This works.
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'''Case 4: the first letter is D'''
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''Subcase 1: the second letter is A''
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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.
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''Subcase 2: the second letter is B''
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The third letter cannot be A or C, so this doesn't work.
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Summing the cases, there are two that work: <math>BDAC</math> and <math>CADB</math> <math>\Longrightarrow \boxed{\textbf{(C)}\ 2}</math>.
 +
 +
~JH. L
 +
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/hTcv8lbvs6o
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
 +
 +
== Video Solution ==
 +
https://youtu.be/0W3VmFp55cM?t=1055
 +
 +
~ pi_is_3.14
 +
 +
== Video Solution ==
 +
https://youtu.be/3MiGotKnC_U?t=1509
 +
 +
~ ThePuzzlr
 +
 +
==Video Solution==
 +
https://youtu.be/8sTQIX4YJ6s
 +
 +
~savannahsolver
  
 
==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}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 18:17, 10 August 2023

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)

https://youtu.be/hTcv8lbvs6o

~Education, the Study of Everything



Video Solution

https://youtu.be/0W3VmFp55cM?t=1055

~ pi_is_3.14

Video Solution

https://youtu.be/3MiGotKnC_U?t=1509

~ ThePuzzlr

Video Solution

https://youtu.be/8sTQIX4YJ6s

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AMC 10 Problems and Solutions

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