Difference between revisions of "2018 AMC 8 Problems/Problem 11"

(Solution)
(Problem 11)
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<math>\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}</math>
 
<math>\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}</math>
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==Solution==
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There are a total of 6! ways to arrange the kids.
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Abby and Bridget can sit in 3 ways if they are adjacent in the same column:
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<cmath>\begin{eqnarray*}
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\text{A}&\quad\text{X}\quad&\text{X} \\
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\text{B}&\quad\text{X}\quad&\text{X}
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\end{eqnarray*}</cmath>
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 +
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<cmath>\begin{eqnarray*}
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\text{X}&\quad\text{A}\quad&\text{X} \\
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\text{X}&\quad\text{B}\quad&\text{X}
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\end{eqnarray*}</cmath>
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 +
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<cmath>\begin{eqnarray*}
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\text{X}&\quad\text{X}\quad&\text{A} \\
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\text{X}&\quad\text{X}\quad&\text{B}
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\end{eqnarray*}</cmath>
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For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in <math>4!</math> ways which results in a total of <math>3 \times 2 \times 4!</math> ways to arrange them.
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By the same logic, there are 4 ways for Abby and Bridget to placed if they are adjacent in the same row, they can swap seats, and the other <math>4</math> people can be arranged in <math>4!</math> ways, for a total of <math>4 \times 2 \times 4!</math> ways to arrange them.
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We sum the 2 possibilities up to get <math>\frac{14*4!}{6!}=\boxed{\frac{7}{15}}</math> or <math>\textbf{(D)}</math> - song2sons
 
{{AMC8 box|year=2018|num-b=10|num-a=12}}
 
{{AMC8 box|year=2018|num-b=10|num-a=12}}

Revision as of 17:27, 21 November 2018

Problem 11

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. \begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{X} \\ \text{X}&\quad\text{X}\quad&\text{X}  \end{eqnarray*} If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$

Solution

There are a total of 6! ways to arrange the kids.

Abby and Bridget can sit in 3 ways if they are adjacent in the same column: \begin{eqnarray*} \text{A}&\quad\text{X}\quad&\text{X} \\ \text{B}&\quad\text{X}\quad&\text{X}  \end{eqnarray*}


\begin{eqnarray*} \text{X}&\quad\text{A}\quad&\text{X} \\ \text{X}&\quad\text{B}\quad&\text{X}  \end{eqnarray*}


\begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{A} \\ \text{X}&\quad\text{X}\quad&\text{B}  \end{eqnarray*}


For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in $4!$ ways which results in a total of $3 \times 2 \times 4!$ ways to arrange them.


By the same logic, there are 4 ways for Abby and Bridget to placed if they are adjacent in the same row, they can swap seats, and the other $4$ people can be arranged in $4!$ ways, for a total of $4 \times 2 \times 4!$ ways to arrange them.


We sum the 2 possibilities up to get $\frac{14*4!}{6!}=\boxed{\frac{7}{15}}$ or $\textbf{(D)}$ - song2sons

2018 AMC 8 (ProblemsAnswer KeyResources)
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Problem 10
Followed by
Problem 12
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