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2015 AMC 12A Problem 17

Problem 17 of 25IntermediateCounting & Probability

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?

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Solution

There are 28=2562^8 = 256 equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of 88 seats, grouped by how many people stand. The number of ways to choose kk non-adjacent seats from a circle of nn is nnk(nkk).\dfrac{n}{n-k}\dbinom{n-k}{k}. For n=8n = 8 this gives 1,  8,  20,  16,  21,\; 8,\; 20,\; 16,\; 2 for k=0,1,2,3,4,k = 0,1,2,3,4, and more than 44 standers is impossible without an adjacency. The total is 1+8+20+16+2=47,1 + 8 + 20 + 16 + 2 = 47, so the probability is 47256.\dfrac{47}{256}. Thus, the correct answer is A.

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Concepts: circular arrangements · basic probability · casework

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.