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| − | ==Problem==
| + | #REDIRECT[[2014 AMC 10A Problems/Problem 4]] |
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| − | 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?
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| − | <math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}}\ 5\qquad\textbf{(E)}\ 6</math>
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| − | ==Solution==
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| − | There are 24 possible arrangements of the houses. The number of ways with the blue house next to the yellow house is <math>3! \cdot 2!=12</math>, as we can consider the arrangements of O, (RB), and Y. Thus there are <math>24-12</math> arrangements with the blue and yellow houses non-adjacent.
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| − | Exactly half of these have the orange house before the red house by symmetry, and exactly half of those have the blue house before the yellow house (also by symmetry), so our answer is <math>12 \cdot \frac{1}{2} \cdot \frac{1}{2}=3</math>.
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| − | (Solution by BOGTRO)
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