Difference between revisions of "2015 AMC 10A Problems/Problem 10"
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− | ==Solution 2 ( | + | ==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. | ||
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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− | == Video Solution | + | ==Video Solution (CREATIVE THINKING)== |
+ | https://youtu.be/hTcv8lbvs6o | ||
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+ | ~Education, the Study of Everything | ||
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+ | == Video Solution == | ||
https://youtu.be/0W3VmFp55cM?t=1055 | https://youtu.be/0W3VmFp55cM?t=1055 | ||
Latest revision as of 18:17, 10 August 2023
Contents
[hide]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:
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.