2015 AMC 12A Problem 17

Question

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?

\dfrac{47}{256}

\dfrac{3}{16}

\dfrac{49}{256}

\dfrac{25}{128}

\dfrac{51}{256}

Answer: A

Concepts:circular arrangements basic probability casework

Difficulty rating: 1910

Small hint:

There are

2^8 = 256

equally likely outcomes; count those with no two adjacent heads

Big hint:

Group by the number of people standing

(0,1,2,3,4);

more than

4

forces an adjacency

Solution:

There are

2^8 = 256

equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of

8

seats, grouped by how many people stand.

The number of ways to choose

k

non-adjacent seats from a circle of

n

is

\dfrac{n}{n-k}\dbinom{n-k}{k}.

For

n = 8

this gives

1,\; 8,\; 20,\; 16,\; 2

for

k = 0,1,2,3,4,

and more than

4

standers is impossible without an adjacency.

The total is

1 + 8 + 20 + 16 + 2 = 47,

so the probability is

\dfrac{47}{256}.

Thus, the correct answer is A.