Difference between revisions of "2011 AMC 8 Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | If we designate | + | 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 <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== | ==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> | 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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | Three people can be seated opposite of Angie (Bridget, Carlos, or Diego). Only one of these is Carlos. Thus, the answer is <math>\boxed{\textbf{(B) }\frac{1}{3}}</math>. | ||
+ | - @scthecool | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/SpTtrFmDAK4 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=11|num-a=13}} | {{AMC8 box|year=2011|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:00, 18 November 2024
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 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 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
Solution 3
Three people can be seated opposite of Angie (Bridget, Carlos, or Diego). Only one of these is Carlos. Thus, the answer is . - @scthecool
Video Solution
https://www.youtube.com/watch?v=mYn6tNxrWBU
Video Solution by WhyMath
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.