Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem 11)
(→‎Solution 2)
 
(37 intermediate revisions by 24 users not shown)
Line 1: Line 1:
−
==Problem 11==
+
==Problem==
−
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.
+
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.  
−
<math>\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*}</math>
+
\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>
−
<math>{{AMC8 box|year=2018|num-b=10|num-a=12}}</math>
+
 
 +
==Solution 1==
 +
 
 +
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:
 +
<cmath>\begin{eqnarray*}
 +
\text{A}&\quad\text{X}\quad&\text{X} \\
 +
\text{B}&\quad\text{X}\quad&\text{X}
 +
\end{eqnarray*}</cmath>
 +
 
 +
 
 +
<cmath>\begin{eqnarray*}
 +
\text{X}&\quad\text{A}\quad&\text{X} \\
 +
\text{X}&\quad\text{B}\quad&\text{X}
 +
\end{eqnarray*}</cmath>
 +
 
 +
 
 +
<cmath>\begin{eqnarray*}
 +
\text{X}&\quad\text{X}\quad&\text{A} \\
 +
\text{X}&\quad\text{X}\quad&\text{B}
 +
\end{eqnarray*}</cmath>
 +
 
 +
 
 +
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.
 +
 
 +
 
 +
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>.
 +
 
 +
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 <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==
 +
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
 +
 
 +
==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==
 +
{{AMC8 box|year=2018|num-b=10|num-a=12}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 08:08, 10 April 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_11&oldid=218052"