2015 AMC 10A Problems/Problem 22

Problem 22

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?

$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$

Solution

We will count how many valid standing arrangements are there (counting rotations as distinct), and divide by $2^8 = 256$ 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 $\dbinom{8}{2} - 8 = 20$ 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 $\dfrac{47}{256}$ $\textbf{(A)}$ .

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2015 AMC 10A Problems/Problem 22

Problem 22

Solution