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2014 AMC 10A Problems/Problem 4

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 (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 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.

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