Difference between revisions of "2011 AMC 8 Problems/Problem 12"
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<math> \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 </math> | <math> \textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
If we designate Angie to be on a certain side, then all placements of the other people can be considered unique. There are then <math>3!=6</math> total seating arrangements. If Carlos is across from Angie, there are only <math>2!=2.</math> ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is <math>\frac26=\boxed{\textbf{(B)}\ \frac13}</math> | If we designate Angie to be on a certain side, then all placements of the other people can be considered unique. There are then <math>3!=6</math> total seating arrangements. If Carlos is across from Angie, there are only <math>2!=2.</math> ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is <math>\frac26=\boxed{\textbf{(B)}\ \frac13}</math> | ||
| + | |||
| + | ==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 <math>\boxed{\textbf{(B) }\frac{1}{3}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=11|num-a=13}} | {{AMC8 box|year=2011|num-b=11|num-a=13}} | ||
Revision as of 21:53, 12 November 2012
Contents
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?
Solution 1
If we designate Angie to be on a certain side, then all placements of the other people can be considered unique. There are then 
total seating arrangements. If Carlos is across from Angie, there are only 
ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is 
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 
See Also
| 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 | ||
