Difference between revisions of "2018 AMC 8 Problems/Problem 11"
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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== | ||
+ | |||
+ | 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 17:27, 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. 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?
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:
For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in ways which results in a total of 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 people can be arranged in ways, for a total of ways to arrange them.
We sum the 2 possibilities up to get or - song2sons
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AJHSME/AMC 8 Problems and Solutions |