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
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:
![\[1 + 1 = \boxed{\textbf{(C)}\ 2}.\]](http://latex.artofproblemsolving.com/f/4/1/f41291c0414c639626de55d3be3727b6fa413201.png)
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. 
