Difference between revisions of "2014 AMC 10A Problems/Problem 4"
(→Solution) |
Flamedragon (talk | contribs) (→See Also) |
||
Line 34: | Line 34: | ||
{{AMC10 box|year=2014|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2014|ab=A|num-b=3|num-a=5}} | ||
+ | {{AMC12 box|year=2014|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:37, 8 February 2014
- The following problem is from both the 2014 AMC 12A #3 and 2014 AMC 10A #4, so both problems redirect to this page.
Contents
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 1
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
Solution 2
There are 24 possible arrangements of the houses. The number of ways with the blue house next to the yellow house is , as we can consider the arrangements of O, (RB), and Y. Thus there are arrangements with the blue and yellow houses non-adjacent.
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 .
(Solution by BOGTRO)
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 |
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.