Difference between revisions of "2015 AMC 10A Problems/Problem 10"
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<cmath>1 + 1 = \boxed{\textbf{(C)}\ 2}.</cmath> | <cmath>1 + 1 = \boxed{\textbf{(C)}\ 2}.</cmath> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/8sTQIX4YJ6s | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 19:52, 17 October 2020
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
The first thing one would want to do is see a possible value 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:
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.