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

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==Problem 11==
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==Problem==
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.  
 
<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}
 
\text{X}&\quad\text{X}\quad&\text{X} \\
 
\text{X}&\quad\text{X}\quad&\text{X} \\
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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>
  
==Solution==
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==Solution 1==
  
 
There are a total of <math>6!</math> ways to arrange the kids.
 
There are a total of <math>6!</math> ways to arrange the kids.
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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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By the same logic, there are 4 ways for Abby and Bridget to be 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.
  
  
 
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>.
 
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>.
  
A more simplistic way to do this is to consider the probability Bridget is adjacent for each of the 6 possible locations for Abby.  If Abby is in any of the 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 (4/6)*(2/5) + (2/6)*(3/5) = 7/15.
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If you got stuck on this problem, refer to AOPS Probability and Combinations
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~Nivaar
  
 
==Solution 2==
 
==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 <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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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>
  
 
==Solution 3==
 
==Solution 3==
The total number of ways is n(S) = C(6,2) = 15, if we treat Abby and Bridget as a pair and forget the others.
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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.
 
The total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns)
 
The total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns)
Therefore, P(Abby and Bridget sitting adjacent) is 7/15 (C)
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Therefore, P(Abby and Bridget sitting adjacent) is <math>\boxed{\textbf{(C) }\frac{7}{15}}</math>.
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==Solution 4==
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We can split the seating into two separate cases: if Abby is sitting on the corners, and if Abby isn't. If Abby is sitting in the corners, there is a <math>\frac{2}{5}</math> chance Bridget is sitting next to Abby, so there is a <math>\frac{2}{5} \cdot \frac{4}{6} = \frac{4}{15}</math> chance for the first case. Meanwhile, if Abby is sitting in the middle row, there is a <math>\frac{3}{5}</math> chance Bridget is sitting next to Abby, so there is a <math>\frac{3}{5} \cdot \frac{2}{6} = \frac{1}{5}</math> chance for the second case. Therefore, P(Abby and Bridget are sitting adjacent to each other) is <math>\frac{4}{15} + \frac{1}{5} = \boxed{\frac{7}{15}}</math> , or <math>\boxed{\textbf{C}}</math>. ~strongstephen
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==Video Solution (CREATIVE ANALYSIS!!!)==
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https://youtu.be/sZhsVX4xIgg
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/YNH7IwMSsh0
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 +
https://youtu.be/EMe9rve8wI0
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~savannahsolver
  
 
==See also==
 
==See also==

Latest revision as of 12:26, 23 January 2024

Problem

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 be 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 $\textbf{(C)}$.

If you got stuck on this problem, refer to AOPS Probability and Combinations

~Nivaar

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 $\boxed{\textbf{(C) }\frac{7}{15}}$.

Solution 4

We can split the seating into two separate cases: if Abby is sitting on the corners, and if Abby isn't. If Abby is sitting in the corners, there is a $\frac{2}{5}$ chance Bridget is sitting next to Abby, so there is a $\frac{2}{5} \cdot \frac{4}{6} = \frac{4}{15}$ chance for the first case. Meanwhile, if Abby is sitting in the middle row, there is a $\frac{3}{5}$ chance Bridget is sitting next to Abby, so there is a $\frac{3}{5} \cdot \frac{2}{6} = \frac{1}{5}$ chance for the second case. Therefore, P(Abby and Bridget are sitting adjacent to each other) is $\frac{4}{15} + \frac{1}{5} = \boxed{\frac{7}{15}}$ , or $\boxed{\textbf{C}}$. ~strongstephen

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/sZhsVX4xIgg

~Education, the Study of Everything

Video Solution

https://youtu.be/YNH7IwMSsh0

https://youtu.be/EMe9rve8wI0

~savannahsolver

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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