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2024 AMC 8 Problems/Problem 25

Revision as of 14:43, 25 January 2024 by Rhydon516 (talk | contribs) (→‎Solution 2 (Straightforward Casework))

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?

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.

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

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