Difference between revisions of "2015 AMC 10A Problems/Problem 22"

Revision as of 00:04, 4 October 2021

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 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 $5.$

Alternate Case $5:$ Use complementary counting. Total number of ways to choose 3 people from 8 which is $\dbinom{8}{3}$ . Sub-case $1:$ three people are next to each other which is $\dbinom{8}{1}$ . Sub-case $2:$ two people are next to each other and the third person is not $\dbinom{8}{1}$ $\dbinom{4}{1}$ . This yields $\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16$

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 sticks and stones aka balls and urns.

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 (Recursion)

Because the denominator is always $256$ , we can count the numerator with a recursion. Define $a_n$ and $b_n$ as the number of satisfying arrangements on a circle and row with $n$ seats, respectively. Then, observe that $a_n=b_{n-1}+b_{n-3}$ $b_n=b_{n-1}+b_{n-2}$ (From casework on whether the nth position is $H,T$ ). Because $b_1=2, b_2=3$ , we can continue to write out each $b_i$ and hence find $a_8=47$ . Our answer is then $\boxed{\textbf{(A) } \dfrac{47}{256}}$ .

Solution 4 (Recursion)

We know that the denominator of the probability is $2^8=256$ . So now we only have to calculate the numerator, which is the number of arrangements for $8$ people at a round table without $2$ or more neighboring people standing.

Denote $a_n$ as number of arrangements for $n$ people at a round table without $2$ or more neighboring people standing. We can see that $a_2=3$ , $a_3=4$ , we are going to prove $a_n=a_{n-1}+a_{n-2}$

We use $1$ to represent standing people, and $0$ as sitting people. The problem becomes arranging $n$ numbers of $0$ or $1$ around a circle with no consecutive $1$ 's.

We elect one person as the chairman, let him be $p_1$ , the first element in this circular sequence. There are $2$ cases to generate the arrangement of $n$ numbers.

$Case$ $1:$ From the arrangement of $n-1$ numbers, add $1$ more number $p_n$ counter-clockwise next to $p_1$ .

If $p_1=1$ , $p_{n-1}p_np_1$ is $001$ , as the diagram shows:

$[asy] draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=0$", C, SW); [/asy]$

If $p_1=0$ , $p_{n-1}p_np_1$ is $X00$ , $X$ could be $0$ or $1$ , as the diagram shows:

$[asy] draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=X$", C, SW); [/asy]$

$Case$ $2:$ From the arrangement of $n-2$ numbers, add $2$ more numbers $p_{n-1}$ and $p_n$ counter-clockwise next to $p_1$ .

If $p_1=1$ , then $p_{n-2}=0$ , let $p_{n-1}=1, p_n=0$ , $p_{n-1}p_np_1$ is $101$ , as the diagram shows:

$[asy] draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=1$", C, SW); [/asy]$

If $p_1=0$ , then $p_{n-2}$ could be $0$ or $1$ . Let $p_{n-1}=0, p_n=1$ , $p_{n-1}p_np_1$ is $010$ , as the diagram shows:

$[asy] draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=1$", B, NW); label("$p_{n-1}=0$", C, SW); [/asy]$

From the above $2$ cases and the $4$ diagrams, the arrangements of $p_{n-1}p_np_1$ are mutually exclusive collectively exhaustive, so $a_{n}=a_{n-1}+a_{n-2}$

The answer is $\boxed{\textbf{(A) } \dfrac{47}{256}}$

This sequence of numbers is called Lucas Number.

~isabelchen

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=krlnSWWp0I0

@@ Line 57: / Line 57: @@
-<math>Case</math> <math>1:</math> From the arrangement of <math>n-1</math> numbers, add one more number <math>p_n</math> counter-clockwise next to <math>p_1</math>.
+<math>Case</math> <math>1:</math> From the arrangement of <math>n-1</math> numbers, add <math>1</math> more number <math>p_n</math> counter-clockwise next to <math>p_1</math>.
 If <math>p_1=1</math>, <math>p_{n-1}p_np_1</math> is <math>001</math>, as the diagram shows:
@@ Line 86: / Line 86: @@
-<math>Case</math> <math>2:</math> From the arrangement of <math>n-2</math> numbers, add 2 more numbers <math>p_{n-1}</math> and <math>p_n</math> counter-clockwise next to <math>p_1</math>.
+<math>Case</math> <math>2:</math> From the arrangement of <math>n-2</math> numbers, add <math>2</math> more numbers <math>p_{n-1}</math> and <math>p_n</math> counter-clockwise next to <math>p_1</math>.
 If <math>p_1=1</math>, then <math>p_{n-2}=0</math>, let <math>p_{n-1}=1, p_n=0</math>, <math>p_{n-1}p_np_1</math> is <math>101</math>, as the diagram shows:

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 10 Problems and Solutions

2015 AMC 12A (Problems • Answer Key • Resources)
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