During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# 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

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

Case 1: $Y$ is the $3$rd house.

The only possible arrangement is $B-O-Y-R$

Case 2: $Y$ is the last house.

There are two possible ways:

$B-O-R-Y$ and

$O-B-R-Y$ 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 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 $12 \cdot \frac{1}{2} \cdot \frac{1}{2}= \boxed{\textbf{(B)} 3}$

~savannahsolver