2015 AMC 10A Problems/Problem 22

The following problem is from both the 2015 AMC 12A #17 and 2015 AMC 10A #22, so both problems redirect to this page.

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?

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

Solution

Solution 1

We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. We casework on how many people are 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 $\dfrac{8 \cdot 6}{3} = 16$ arrangements for Case $5.$

Summing gives $1 + 8 + 20 + 2 + 16 = 47,$ and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 2

We will count how many valid standing arrangements there are counting rotations as distinct and divide by $256$ at the end. Line up all $8$ 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 $2$ 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 ball and urn.

If there are $4$ standing, there are ${4 \choose 4}=1$ ways to place them. For $3,$ there are ${3+2 \choose 3}=10$ ways. etc. Summing, we get ${4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34$ ways.

Now we consider that the far right person can be standing as well, so we have ${3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13$ ways

Together we have $34+13=47$ , and so our probability is $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 3

We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by $2^8 = 256$ at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only $7$ people. If they stand, we count the arrangements with $6$ instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with $1$ person there are two ways and with $2$ people there are three ways. Carrying out the Fibonacci recursion until we get to $8$ people, we find there are $55$ standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for $4$ people to stand in a line, which is $8$ from our sequence. Therefore our probability is $\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}$

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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