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

Video Solution

https://www.youtube.com/watch?v=mYn6tNxrWBU

Video Solution by WhyMath

https://youtu.be/SpTtrFmDAK4

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search