2015 AMC 10A Problems/Problem 10
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 .
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 Formatting seems a bit weird, if anyone is familiar with AoPS formatting please fix it, thanks :)
Video Solution by OmegaLearn
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