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

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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>.
  
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A more simplistic way to do this is to consider the probability Bridget is adjacent to each of the 6 possible locations for Abby.  If Abby is in any of 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 <math>\dfrac{4}{6} * \dfrac{2}{5} + \dfrac{2}{6}* \dfrac{3}{5} = \boxed{\dfrac{7}{15}}.</math>
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A more simplistic way to do this is to consider the probability Bridget is adjacent to each of the 6 possible locations for Abby.  If Abby is in any of 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 <math>\dfrac{4}{6} \cdot \dfrac{2}{5} + \dfrac{2}{6} \cdot \dfrac{3}{5} = \boxed{\dfrac{7}{15}}.</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:05, 25 October 2020

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 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 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 $\textbf{(C)}$.

A more simplistic way to do this is to consider the probability Bridget is adjacent to each of the 6 possible locations for Abby. If Abby is in any of 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 $\dfrac{4}{6} \cdot \dfrac{2}{5} + \dfrac{2}{6} \cdot \dfrac{3}{5} = \boxed{\dfrac{7}{15}}.$

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

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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