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

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==Solution 2 (casework)==
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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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'''Case 1: the first letter is A'''
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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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''Subcase 1: the second 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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The next letter must either be B or D, both of which do not satisfy the conditions.
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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 D''
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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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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.
  
 
Summing the cases, there are two that work: <math>BDAC</math> and <math>CADB</math> <math>\Longrightarrow \boxed{\textbf{(C)}\ 2}</math>.
 
Summing the cases, there are two that work: <math>BDAC</math> and <math>CADB</math> <math>\Longrightarrow \boxed{\textbf{(C)}\ 2}</math>.
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~JH. L
 
~JH. L
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Formatting seems a bit weird, if anyone is familiar with AoPS formatting please fix it, thanks :)
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/hTcv8lbvs6o
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~Education, the Study of Everything
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== Video Solution ==
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https://youtu.be/0W3VmFp55cM?t=1055
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~ pi_is_3.14
  
 
== Video Solution ==
 
== Video Solution ==

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
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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