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2014 AMC 10A Problems/Problem 4

Revision as of 19:18, 7 February 2014 by TheMaskedMagician (talk | contribs)
The following problem is from both the 2014 AMC 12A #3 and 2014 AMC 10A #4, so both problems redirect to this page.

Problem

Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}}\ 5\qquad\textbf{(E)}\ 6$ (Error compiling LaTeX. ! Extra }, or forgotten $.)


Solution

Attack this problem with very simple casework. The only possible locations for the yellow house $(Y)$ is the $3$rd house and the last house.

Case 1: $Y$ is the $3$rd house.

The only possible arrangement is $B-O-Y-R$

Case 2: $Y$ is the last house.

There are two possible ways:

$B-O-R-Y$ and

$O-B-R-Y$ so our answer is $\boxed{\textbf{(B)} 3}$

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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

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