2024 AMC 8 Problems/Problem 25

1 Problem
2 Video Solution (4 minutes) by MegaMath
3 Solution 1 (Complementary Counting Casework)
4 Solution 2 (Straightforward Casework)
5 Solution 3 (Complementary Casework on Middle Seats)
6 Video Solution 1 by Math-X (First understand the problem!!!)
7 Video Solution 2 by OmegaLearn.org
8 Video Solution 3 by SpreadTheMathLove
9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
10 See Also

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

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

2024 AMC 8 (Problems • Answer Key • Resources)
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