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"

(→‎Solution)
(→‎Problem 11)
Line 8: Line 8:
  
 
<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==
 +
 +
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:
 +
<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 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{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 18:27, 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 (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_11&oldid=98838"