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2018 AMC 8 Problems/Problem 11: Difference between revisions

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==Problem 11==
==Problem==
Abby, Ayush, and four of their classmates will be seated in two rows of three for a group picture, as shown. Ayush has to be next to Abby, because he is stalking her.
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} \\
\text{X}&\quad\text{X}\quad&\text{X}  
\text{X}&\quad\text{X}\quad&\text{X}  
\end{eqnarray*}</cmath>
\end{eqnarray*}</cmath>
If the seating positions are assigned randomly, what is the probability that Abby and Ayush 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==
==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.


Abby and Bridget can sit in 3 ways if they are adjacent in the same column:
Abby and Bridget can sit in 3 ways if they are adjacent in the same column:
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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.
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.
If you got stuck on this problem, refer to AOPS Probability and Combinations.
 
~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>
We can ignore other students, and treat Abby and Bridget as indistinguishable (since we only care about adjacency, not their order). Thus, the total number of ways is <math>n(S) = _{6}C_{2} = 15</math> .
In one row, they can be adjacent 2 ways:  <math>2 \cdot 2 rows = 4</math>.  
In one column, they can only be adjacent 1 way: <math>1 \cdot 3 cols = 3</math>.  
Add these cases <math>4+3=7</math>, and therefore, P(Abby and Bridget sitting adjacent) is <math>\boxed{\textbf{(C) }\frac{7}{15}}</math>.


==Solution 3==
==Solution 3==
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.
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
The total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns)
 
Therefore, P(Abby and Bridget sitting adjacent) is <math>\boxed{\frac{7}{15}{} (C)}</math>
==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==
==See also==
Line 54: Line 69:


{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Combinatorics Problems]]
[[Category:Introductory Probability Problems]]

Latest revision as of 12:42, 6 June 2025

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 other students, and treat Abby and Bridget as indistinguishable (since we only care about adjacency, not their order). Thus, the total number of ways is $n(S) = _{6}C_{2} = 15$ . In one row, they can be adjacent 2 ways: $2 \cdot 2 rows = 4$. In one column, they can only be adjacent 1 way: $1 \cdot 3 cols = 3$. Add these cases $4+3=7$, and therefore, P(Abby and Bridget sitting adjacent) is $\boxed{\textbf{(C) }\frac{7}{15}}$.

Solution 3

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 (Problems • Answer Key • Resources)
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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