Difference between revisions of "2015 AMC 10A Problems/Problem 10"
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Therefore, our answer must be <math>\boxed{\textbf{(C) }2}</math>. | Therefore, our answer must be <math>\boxed{\textbf{(C) }2}</math>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:06, 4 February 2015
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 
.
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 3\qquad\textbf{(E)}\ 4$ (Error compiling LaTeX. Unknown error_msg)
Solution
Observe that we can't begin a rearrangement with either 
or 
, leaving 
and 
, respectively.
Starting with 
, there is only one rearrangement: 
. Similarly, there is only one rearrangement when we start with 
: 
.
Therefore, our answer must be 
.
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. 
