2018 AMC 8 Problems/Problem 11: Difference between revisions

Revision as of 08:03, 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 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

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 ways * 2 rows = 4 cases. In one column, they can only be adjacent 1 way: 1 way * 3 rows = 3 cases. Add these cases 4+3=7, and therefore, P(Abby and Bridget sitting adjacent) is $\boxed{\textbf{(C) }\frac{7}{15}}$ .

Solution 4

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

@@ Line 48: / Line 48: @@
 ==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 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> .
-The total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns)
+In one row, they can be adjacent 2 ways:  2 ways * 2 rows = 4 cases.
-Therefore, P(Abby and Bridget sitting adjacent) is <math>\boxed{\textbf{(C) }\frac{7}{15}}</math>.
+In one column, they can only be adjacent 1 way: 1 way * 3 rows = 3 cases.
+Add these cases 4+3=7, and therefore, P(Abby and Bridget sitting adjacent) is <math>\boxed{\textbf{(C) }\frac{7}{15}}</math>.
 ==Solution 4==

2018 AMC 8 (Problems • Answer Key • Resources)
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