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

2024 AMC 8 Problems/Problem 25

Revision as of 18:10, 28 January 2024 by Azc1027 (talk | contribs) (→‎Solution 5 (PIE))

Problem

A small airplane has $4$ rows of seats with $3$ seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?

$\textbf{(A)} \frac{8}{15}\qquad\textbf{(B)} \frac{32}{55}\qquad\textbf{(C) } \frac{20}{33}\qquad\textbf{(D) } \frac{34}{55}\qquad\textbf{(E) } \frac{8}{11}$

Video Solution (4 minutes) by MegaMath

https://www.youtube.com/watch?v=DgQsljPaE5Y

Solution 1 (Complementary Counting Casework)

Suppose the passengers are indistinguishable. There are $\binom{12}{8} = 495$ total ways to distribute the passengers. We proceed with complementary counting, and instead, will count the number of passenger arrangements such that the couple cannot sit anywhere. Consider the partitions of $8$ among the rows of $3$ seats, to make our lives easier, assume they are non-increasing. We have $(3, 3, 2, 0), (3, 3, 1, 1), (3, 2, 2, 1), (2, 2, 2, 2)$.


For the first partition, clearly the couple will always be able to sit in the row with $0$ occupied seats, so we have $0$ cases here.


For the second partition, there are $\frac{4!}{2!2!} = 6$ ways to permute the partition. Now the rows with exactly $1$ passenger must be in the middle, so this case generates $6$ cases.


For the third partition, there are $\frac{4!}{2!} = 12$ ways to permute the partition. For rows with $2$ passengers, there are $\binom{3}{2} = 3$ ways to arrange them in the row so that the couple cannot sit there. The row with $1$ passenger must be in the middle. We obtain $12 \cdot 3^2 = 108$ cases.


For the fourth partition, there is $1$ way to permute the partition. As said before, rows with $2$ passengers can be arranged in $3$ ways, so we obtain $3^4 = 81$ cases.


Collectively, we obtain a grand total of $6 + 108 + 81 = 195$ cases. The final probability is $1 - \frac{195}{495} = \boxed{\textbf{(C)}~\frac{20}{33}}$.

~blueprimes [1]

Solution 2 (Straightforward Casework)

Suppose the passengers are indistinguishable.

What this question is asking, is really, if 4 empty seats are placed, what is the probability that there are 2 adjacent seats open.

We proceed by casework.


Case 1: There is exactly one pair of open seats. Then the other seat in that row must be occupied. The other two empty seats are distributed across the remaining $3$ rows without being adjacent, which is $\binom{9}{2}-6=30$ cases per pair of open seats for $30\cdot8=240$ total cases.


Case 2: There is one row of open seats. $4$ ways to choose the row and $9$ to choose the final empty seat for $4\cdot9=36$ cases.


Case 3: There are $2$ independent pairs of open seats. Choose the $2$ rows, then the placement of each pair within each row for $\binom{4}{2}\cdot2^2=24$ cases.


In total, we get $240+36+24=300$ cases total for a probability of \[\frac{300}{\binom{12}{4}}=\frac{300}{495}=\boxed{\mathbf{(C)}~\frac{20}{33}}\]

~rhydon516

Solution 3 (Complementary Casework on Middle Seats)

We notice that if we have a middle seat in a row, then the couple cannot sit in that row. So, we perform complementary casework.

Case 1: Four people sitting in middle seats.

In this case, there are 4 people left to order, and 8 seats. This gives $\dbinom{8}{4}$ total combinations for this case.

Case 2: Three people sitting in middle seats.

In this case, there are $\dbinom{4}{3}$ ways to permute the rows in which the middle seat is occupied. For the row in which the people do not occupy the middle row, we must have two people sitting at the ends of the rows to guarantee the couple cannot sit there. So, for the rest of the 3 people, there are 6 possible seats. So, there are $\dbinom{4}{3} \cdot \dbinom{6}{3}$ total combinations.

Case 3: Two people sitting in middle seats.

In this case, there are $\dbinom{4}{2}$ ways to permute the rows in which the middle seat is occupied. For the rows in which the people do not occupy the middle row, we must have two people sitting at the ends of the rows to guarantee the couple cannot sit there. So, for the rest of the 2 people, there are 4 possible seats. So, there are $\dbinom{4}{2} \cdot \dbinom{4}{2}$ total combinations.

Case 4: One person sitting in a middle seat

In this case, there are $\dbinom{4}{1}$ ways to permute the rows in which the middle seat is occupied. For the rows in which the people do not occupy the middle row, we must have two people sitting at the ends of the rows to guarantee the couple cannot sit there. So, for the rest of the last person, there are 2 possible seats. So, there are $\dbinom{4}{1} \cdot \dbinom{2}{1}$ total combinations.

Case 5: Zero people sitting in a middle seat

In this case, we must have every person sitting at the ends of the seats. So, there is only 1 combination.

In total, we have

\[\dbinom{8}{4} + \dbinom{4}{3} \cdot \dbinom{6}{3} + \dbinom{4}{2} \cdot \dbinom{4}{2} + \dbinom{4}{1} \cdot \dbinom{2}{1} +1\]

combinations, or 195 combinations. The final step is to find the total amount of combinations without restrictions. This is simply $\dbinom{12}{4} = 495$. So, finally employing complementary counting, we have that the probability that there will be 2 adjacent seats for the couple is

\[1 - \dfrac{195}{495} = \dfrac{20}{33}.\]

~NTfish

Solution 4 (Permutations)

There are $12\cdot 11 = 132$ for two people (the married couple) to be seated. This will be our denominator.

There are $8$ pairs of seats that are next to each other in the diagram ($2$ per row; left-middle and middle-right). This will be our numerator.

Since there are $8+2=10$ total people on the plane, we should multiply our numerator by that to account for all ways the 10 people could be seated (e.x. the husband and the wife could be switched around and it would still work, same applies to the other passengers)

Therefore, our numerator is $8 \cdot 10 = 80$.

This creates the fraction $\frac{80}{132}$, which simplifies to \[\boxed{\frac{20}{33}}.\]

- Siddharth Mirchandani (svm2020), John Adams Middle School

Solution 5 (PIE)

There are ${12\choose{8}}$ total ways to choose seats for the first $8$ passengers. Now, we have to count the number of ways to choose seats such that $2$ adjacent seats are empty. There are $8$ ways to choose the $2$ empty seats and ${10\choose{8}}$ ways to choose the seats for the other passengers, so there are ${8 \cdot {10\choose{8}}}$ ways. However, this overcounts the ways that have $2$ sets of consecutive empty seats. There can either be 2 separate pairs of empty seats, or $1$ row of empty seats. There are $\frac{8\cdot 6}{2}$ ways to choose 2 separate pairs of empty seats and the remaining $8$ seats are for the remaining $8$ passengers. There are $4 \cdot {9\choose{8}}$ ways to have an empty row of seats, since there are $4$ ways to choose that row and ${9\choose{8}}$ ways to seat the remaining passengers. Hence, the total number of ways to choose the seats for the other passengers such that there are 2 empty adjacent seats is \[8 \cdot {10\choose{8}} - \frac{8\cdot 6}{2} - 4 \cdot {9\choose{8}} = 360-24-36=300.\] Hence, the total probability is $\frac{300}{{12\choose{8}}} = \frac{300}{11\cdot5\cdot9} = \frac{20}{33}.$ So the answer is $\boxed{\frac{20}{33}}.$ ~azc1027

Video Solution 1 by Math-X (First understand the problem!!!)

https://www.youtube.com/watch?v=tws4rcd1ykc&t=35s

~Math-X

Video Solution 2 by OmegaLearn.org

https://youtu.be/WYxfsShInyM

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=ArN4qVlBDTM

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=bxZHXPMcsGI

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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=2024_AMC_8_Problems/Problem_25&oldid=213612"