Difference between revisions of "2011 AMC 8 Problems/Problem 12"
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==Solution== | ==Solution== | ||
| + | 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> | ||
==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 20:35, 25 November 2011
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
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 
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 | ||
