2014 AMC 10A Problems/Problem 4

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

Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or 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 arrangements:

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

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

The answer is $1+2=\boxed{\textbf{(B) } 3}$

Solution 2

There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.

By symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, 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

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.

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