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

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==Problem 11==
 
==Problem 11==
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.
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Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.  
\begin{eqnarray*}
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<cmath>\begin{eqnarray*}
 
\text{X}&\quad\text{X}\quad&\text{X} \\
 
\text{X}&\quad\text{X}\quad&\text{X} \\
 
\text{X}&\quad\text{X}\quad&\text{X}  
 
\text{X}&\quad\text{X}\quad&\text{X}  
\end{eqnarray*}
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\end{eqnarray*}</cmath>
 
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?
 
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?
  
 
<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 1==
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There are a total of <math>6!</math> 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{(3\cdot2)\cdot4!+(4\cdot2)\cdot4!}{6!} = \frac{14\cdot4!}{6!}=\boxed{\frac{7}{15}}</math> or <math>\textbf{(C)}</math>.
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A more simplistic way to do this is to consider the probability Bridget is adjacent to each of the 6 possible locations for Abby.  If Abby is in any of the corners, the chance that Bridget is adjacent is 2/5 because 2 of 5 possible locations for Bridget is an adjacent location.  If Abby is in either of the two middle locations, the chance that Bridget is adjacent is 3/5 because 3 of 5 locations for Bridget is an adjacent location.  So the total probability they are adjacent is <math>\dfrac{4}{6} \cdot \dfrac{2}{5} + \dfrac{2}{6} \cdot \dfrac{3}{5} = \boxed{\dfrac{7}{15}}.</math>
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==Solution 2==
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We can ignore about the 4 other classmates because they aren't relevant. We can treat Abby and Bridget as a pair, so there are <math>{6 \choose 2}=15</math> total ways to seat them. If they sit in the same row, there are <math>2\cdot2=4</math> ways to seat them. If they sit in the same column, there are <math>3</math> ways to seat them. Thus our answer is <math>\frac{4+3}{15} = \boxed{\textbf{(C) }\frac 7{15}}</math>
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==Solution 3==
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The total number of ways is <math>n(S) = _{6}C_{2} = 15</math> , if we treat Abby and Bridget as a pair and distinguishable and forget the others.
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The total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns)
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Therefore, P(Abby and Bridget sitting adjacent) is <math>\boxed{\frac{7}{15}{} (C)}</math>
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==See also==
 
{{AMC8 box|year=2018|num-b=10|num-a=12}}
 
{{AMC8 box|year=2018|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 22:05, 25 October 2020

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 1

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 $\frac{(3\cdot2)\cdot4!+(4\cdot2)\cdot4!}{6!} = \frac{14\cdot4!}{6!}=\boxed{\frac{7}{15}}$ or $\textbf{(C)}$.

A more simplistic way to do this is to consider the probability Bridget is adjacent to each of the 6 possible locations for Abby. If Abby is in any of the corners, the chance that Bridget is adjacent is 2/5 because 2 of 5 possible locations for Bridget is an adjacent location. If Abby is in either of the two middle locations, the chance that Bridget is adjacent is 3/5 because 3 of 5 locations for Bridget is an adjacent location. So the total probability they are adjacent is $\dfrac{4}{6} \cdot \dfrac{2}{5} + \dfrac{2}{6} \cdot \dfrac{3}{5} = \boxed{\dfrac{7}{15}}.$

Solution 2

We can ignore about the 4 other classmates because they aren't relevant. We can treat Abby and Bridget as a pair, so there are ${6 \choose 2}=15$ total ways to seat them. If they sit in the same row, there are $2\cdot2=4$ ways to seat them. If they sit in the same column, there are $3$ ways to seat them. Thus our answer is $\frac{4+3}{15} = \boxed{\textbf{(C) }\frac 7{15}}$

Solution 3

The total number of ways is $n(S) = _{6}C_{2} = 15$ , if we treat Abby and Bridget as a pair and distinguishable and forget the others. The total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns) Therefore, P(Abby and Bridget sitting adjacent) is $\boxed{\frac{7}{15}{} (C)}$

See also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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