Difference between revisions of "2015 AMC 10A Problems/Problem 22"
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Revision as of 21:18, 4 February 2015
- The following problem is from both the 2015 AMC 12A #17 and 2015 AMC 10A #22, so both problems redirect to this page.
Problem
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
Solution
We will count how many valid standing arrangements are there (counting rotations as distinct), and divide by 
at the end. We casework on how many people is standing.
Case 1: 0 people are standing. This yields 1 arrangement.
Case 2: 1 person is standing. This yields 8 arrangements.
Case 3: 2 people are standing. This yields 
arrangements, because the two people cannot be next to each other.
Case 4: 4 people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding 2 possible arrangements.
More difficult is:
Case 5: 3 people are standing. First, choose the location of the first person standing (8 choices). Next, choose 2 of the remaining people in the remaining 5 legal seats to stand, amounting to 6 arrangements considering that these two people cannot stand next to each other. However, we have to divide by 3, because there are 3 ways to choose the first person given any three. This yields 8 * 6 / 3 = 16 arrangements for Case 5.
Summing gives 1 + 8 + 20 + 2 + 16 = 47, and so our probability is 
.
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AMC 10 Problems and Solutions | ||
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
