Difference between revisions of "2014 AMC 10A Problems/Problem 4"
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==Solution== | ==Solution== | ||
+ | Attack this problem with very simple casework. The only possible locations for the yellow house <math>(Y)</math> is the <math>3</math>rd house and the last house. | ||
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+ | Case 1: <math>Y</math> is the <math>3</math>rd house. | ||
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+ | The only possible arrangement is <math>B-O-Y-R</math> | ||
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+ | Case 2: <math>Y</math> is the last house. | ||
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+ | There are two possible ways: | ||
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+ | <math>B-O-R-Y</math> and | ||
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+ | <math>O-B-R-Y</math> so our answer is <math>\boxed{\textbf{(B)} 3}</math> | ||
==See Also== | ==See Also== |
Revision as of 23:54, 6 February 2014
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 is the rd house and the last house.
Case 1: is the rd house.
The only possible arrangement is
Case 2: is the last house.
There are two possible ways:
and
so our answer is
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
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 |
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