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 22:54, 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. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
