Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 10A Problems/Problem 4"

m (Formatting Solution 1)
(Adding another solution)
Line 20: Line 20:
 
<math>\text{B}-\text{O}-\text{R}-\text{Y}</math> and
 
<math>\text{B}-\text{O}-\text{R}-\text{Y}</math> and
  
−
<math>\text{O}-\text{B}-\text{R}-\text{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==
Line 26: Line 26:
 
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.  
 
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 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>
+
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==
 
==Video Solution==
 
https://youtu.be/XR661k7tLCU
 
https://youtu.be/XR661k7tLCU

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?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution 1

Attack this problem with very simple casework. The only possible locations for the yellow house $(\text{Y})$ is the $3^\text{rd}$ house and the last house.

Case 1: $\text{Y}$ is the $3^\text{rd}$ house.

The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$

Case 2: $\text{Y}$ is the last house.

There are two possible ways:

$\text{B}-\text{O}-\text{R}-\text{Y}$ and

$\text{O}-\text{B}-\text{R}-\text{Y}$ so our answer is $\boxed{\textbf{(B) } 3}$

Solution 2

There are 24 possible arrangements of the houses. The number of ways with the blue house next to the yellow house is $3! \cdot 2!=12$, as we can consider the arrangements of (BY), O, and R. Thus there are $24-12$ 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 $12 \cdot \frac{1}{2} \cdot \frac{1}{2}= \boxed{\textbf{(B) } 3}$

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 $2$ cases ($\text{BORY}$, $\text{BOYR}$). If the blue house is second, the orange house must be first and the yellow house should be fourth, leading to $1$ case ($\text{OBRY}$). Therefore, our answer is $\boxed{\textbf{(B) } 3}$.

~MathFun1000

Video Solution

https://youtu.be/XR661k7tLCU

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_4&oldid=161781"