Difference between revisions of "2015 AMC 10A Problems/Problem 22"
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Together we have <math>34+13=47</math>, and so our probability is <math>\boxed{\textbf{(A) } \dfrac{47}{256}}</math>. | Together we have <math>34+13=47</math>, and so our probability is <math>\boxed{\textbf{(A) } \dfrac{47}{256}}</math>. | ||
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===Solution 5=== | ===Solution 5=== |
Revision as of 22:35, 16 December 2020
- The following problem is from both the 2015 AMC 12A #17 and 2015 AMC 10A #22, so both problems redirect to this page.
Contents
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?
Solutions
Solution 2
We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by at the end. We casework on how many people are standing.
Case people are standing. This yields arrangement.
Case person is standing. This yields arrangements.
Case people are standing. This yields arrangements, because the two people cannot be next to each other.
Case people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding possible arrangements.
More difficult is:
Case people are standing. First, choose the location of the first person standing ( choices). Next, choose of the remaining people in the remaining legal seats to stand, amounting to arrangements considering that these two people cannot stand next to each other. However, we have to divide by because there are ways to choose the first person given any three. This yields arrangements for Case
Alternate Case Use complementary counting. Total number of ways to choose 3 people from 8 which is . Sub-case three people are next to each other which is . Sub-case two people are next to each other and the third person is not . This yields
Summing gives and so our probability is .
Solution 3
We will count how many valid standing arrangements there are counting rotations as distinct and divide by at the end. Line up all people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.
If there are standing, there are ways to place them. For there are ways. etc. Summing, we get ways.
Now we consider that the far right person can be standing as well, so we have ways
Together we have , and so our probability is .
Solution 5
We will count the number of valid arrangements and then divide by at the end. We proceed with casework on how many people are standing.
Case people are standing. This yields arrangement.
Case person is standing. This yields arrangements.
Case people are standing. To do this, we imagine having 6 people with tails in a line first. Notate "tails" with . Thus, we have . Now, we look to distribute the 2 's into the 7 gaps made by the 's. We can do this in ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have arrangements.
Case people are standing. Similarly, we imagine 5 's. Thus, we have . We distribute 3 's into the gaps, which can be done ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: ) Thus, we have arrangements.
Case people are standing. This can clearly be done in 2 ways: or . This yields arrangements.
Summing the cases, we get arrangements. Thus, the probability is
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=krlnSWWp0I0
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 |
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