Difference between revisions of "2015 AMC 10A Problems/Problem 10"
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~JH. L | ~JH. L | ||
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| + | ==Solution 3== | ||
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| + | We note that <math>a</math> must be adjacent to <math>c</math> or <math>d</math>, <math>b</math> must be adjacent to <math>d</math>, <math>c</math> must be adjacent to <math>a</math>, and <math>d</math> must be adjacent to <math>a</math> or <math>b</math>. In other words, the way the letters can be connected is <cmath>c-a-d-b.</cmath> Therefore, the only two possible arrangements are <math>cadb</math> and <math>bdac</math>, which gives <math>\boxed{\textbf{(C)}\ 2}</math>. | ||
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==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
Latest revision as of 21:43, 3 November 2025
Contents
Problem
How many rearrangements of 
are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either 
or 
.
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 
, we can only place a 
or 
next to it. Unfortunately, after that step, we can't do too much, since:
is not allowed because of the 
, and 
is not allowed because of the 
.
We get the same problem if we start with a , since a 
will have to end up in the middle, causing it to be adjacent to an or .
If we start with a , the next letter would have to be a , and since we can put an next to it and then a after that, this configuration works. The same approach applies if we start with a .
So the solution must be the two solutions that were allowed, one starting from a and the other with a , giving us:
![\[1 + 1 = \boxed{\textbf{(C)}\ 2}.\]](http://latex.artofproblemsolving.com/f/4/1/f41291c0414c639626de55d3be3727b6fa413201.png)
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: 
and 
.
~JH. L
Solution 3
We note that must be adjacent to or , must be adjacent to , must be adjacent to , and must be adjacent to or . In other words, the way the letters can be connected is ![\[c-a-d-b.\]](http://latex.artofproblemsolving.com/9/9/c/99c684381ac7c78761c0d8fd8bc10e3463e51012.png)
Therefore, the only two possible arrangements are 
and 
, which gives 
.
Video Solution (CREATIVE THINKING)
~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
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
