Difference between revisions of "2014 AMC 10A Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
− | Attack this problem with very simple casework. The only possible locations for the yellow house <math>(Y)</math> is the <math>3</math> | + | Attack this problem with very simple casework. The only possible locations for the yellow house <math>(\text{Y})</math> is the <math>3^\text{rd}</math> house and the last house. |
− | Case 1: <math>Y</math> is the <math>3</math> | + | Case 1: <math>\text{Y}</math> is the <math>3^\text{rd}</math> house. |
− | The only possible arrangement is <math>B-O-Y-R</math> | + | The only possible arrangement is <math>\text{B}-\text{O}-\text{Y}-\text{R}</math> |
− | Case 2: <math>Y</math> is the last house. | + | Case 2: <math>\text{Y}</math> is the last house. |
There are two possible ways: | There are two possible ways: | ||
− | <math>B-O-R-Y</math> and | + | <math>\text{B}-\text{O}-\text{R}-\text{Y}</math> and |
− | <math>O-B-R-Y</math> so our answer is <math>\boxed{\textbf{(B)} 3}</math> | + | <math>\text{O}-\text{B}-\text{R}-\text{Y}</math> so our answer is <math>\boxed{\textbf{(B) } 3}</math> |
==Solution 2== | ==Solution 2== | ||
− | 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 | + | 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 (BY), O, and R. Thus there are <math>24-12</math> arrangements with the blue and yellow houses non-adjacent. |
− | Exactly half of these have the | + | Exactly half of these have the blue house before the yellow house by symmetry, and exactly half of those 6 arrangements have the orange house before the red house (also by symmetry), so our answer is <math>12 \cdot \frac{1}{2} \cdot \frac{1}{2}= \boxed{\textbf{(B) } 3}</math> |
+ | |||
+ | ==Solution 3== | ||
+ | This solution is an alternate to Solution 1. | ||
+ | |||
+ | First, we realize that the blue house is the first or second one. If the blue house is the first, then the orange house must be second, leading to <math>2</math> cases (<math>\text{BORY}</math>, <math>\text{BOYR}</math>). If the blue house is second, the orange house must be first and the yellow house should be fourth, leading to <math>1</math> case (<math>\text{OBRY}</math>). Therefore, our answer is <math>\boxed{\textbf{(B) } 3}</math>. | ||
+ | |||
+ | ~MathFun1000 | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/XR661k7tLCU | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 09:52, 7 September 2021
- The following problem is from both the 2014 AMC 12A #3 and 2014 AMC 10A #4, so both problems redirect to this page.
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?
Solution 1
Attack this problem with very simple casework. The only possible locations for the yellow house is the house and the last house.
Case 1: is the 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 (BY), O, and R. Thus there are arrangements with the blue and yellow houses non-adjacent.
Exactly half of these have the blue house before the yellow house by symmetry, and exactly half of those 6 arrangements have the orange house before the red house (also by symmetry), so our answer is
Solution 3
This solution is an alternate to Solution 1.
First, we realize that the blue house is the first or second one. If the blue house is the first, then the orange house must be second, leading to cases (, ). If the blue house is second, the orange house must be first and the yellow house should be fourth, leading to case (). Therefore, our answer is .
~MathFun1000
Video Solution
~savannahsolver
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.