# Difference between revisions of "2018 AMC 8 Problems/Problem 11"

## Problem 11

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. $\begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{X} \\ \text{X}&\quad\text{X}\quad&\text{X} \end{eqnarray*}$ If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$

## Solution

There are a total of 6! ways to arrange the kids.

Abby and Bridget can sit in 3 ways if they are adjacent in the same column: $\begin{eqnarray*} \text{A}&\quad\text{X}\quad&\text{X} \\ \text{B}&\quad\text{X}\quad&\text{X} \end{eqnarray*}$

$\begin{eqnarray*} \text{X}&\quad\text{A}\quad&\text{X} \\ \text{X}&\quad\text{B}\quad&\text{X} \end{eqnarray*}$

$\begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{A} \\ \text{X}&\quad\text{X}\quad&\text{B} \end{eqnarray*}$

For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in $4!$ ways which results in a total of $3 \times 2 \times 4!$ ways to arrange them.

By the same logic, there are 4 ways for Abby and Bridget to placed if they are adjacent in the same row, they can swap seats, and the other $4$ people can be arranged in $4!$ ways, for a total of $4 \times 2 \times 4!$ ways to arrange them.

We sum the 2 possibilities up to get $\frac{14*4!}{6!}=\boxed{\frac{7}{15}}$ or $\textbf{(D)}$ - song2sons

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