Difference between revisions of "2015 AMC 10A Problems/Problem 10"

(See also: Added See Also section.)
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
Observe that we can't begin a rearrangement with either <math>a</math> or <math>d</math>, leaving <math>bcd</math> and <math>abc</math>, respectively.
 
 
Starting with <math>b</math>, there is only one rearrangement: <math>bdac</math>. Similarly, there is only one rearrangement when we start with <math>c</math>: <math>cadb</math>.
 
 
Therefore, our answer must be <math>\boxed{\textbf{(C) }2}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:25, 21 March 2015

Problem

How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$.

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 3\qquad\textbf{(E)}\ 4$ (Error compiling LaTeX. Unknown error_msg)


Solution

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png