2024 AMC 8

Solution 1 

Suppose the passengers are indistinguishable. There are 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 among the rows of seats, to make our lives easier, assuming they are non-increasing. We have .

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

For the second partition, there are ways to permute the partition. Now the rows with exactly passenger must be in the middle, so this case generates cases.

For the third partition, there are ways to permute the partition. For rows with passengers, there are ways to arrange them in the row so that the couple cannot sit there. The row with passenger must be in the middle. We obtain cases.

For the fourth partition, there is way to permute the partition. As said before, rows with passengers can be arranged in ways, so we obtain cases.

Collectively, we obtain a total of cases. The final probability is .

Solution 2

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 rows without being adjacent, which is cases per pair of open seats for total cases.

Case 2: There is one row of open seats. ways to choose the row and to choose the final empty seat for cases.

Case 3: There are independent pairs of open seats. Choose the rows, then the placement of each pair within each row for cases.

In total, we get cases total for a probability of