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counting and probability senior

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?
(A)
(B)
(C)
(D)
Solution
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 . Alternate: (Quicksolve) - We know that for case 5, there are 8 ways to choose the first person, and 3 ways to choose the first person given any 3 - which means that for the case, there is 8 * something divided by 3. The sum of the other cases is 31/256. Thus, add a multiple of 8 to 31 to get an answer. The options are 31 + 8 = 39, 31 + 16 = 47, 31 + 24 = 55, etc. The only possible answer is 47/256.
Final answer
A