# 2011 AMC 8 Problems/Problem 12

## Problem

Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?

$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$

## Solution 1

If we designate a person to be on a certain side, then all placements of the other people can be considered unique. WLOG, assign Angie to be on the side. There are then $3!=6$ total seating arrangements. If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is $\frac26=\boxed{\textbf{(B)}\ \frac13}$ .

## Solution 2

If we seat Angie first, there would be only one out of three ways Carlos can sit across from Angie. So the final answer is $\boxed{\textbf{(B) }\frac{1}{3}}$

## Solution 3

Three people can be seated opposite of Angie (Bridget, Carlos, or Diego). Only one of these is Carlos. Thus, the answer is $\boxed{\textbf{(B) }\frac{1}{3}}$. - @scthecool