Difference between revisions of "2024 AMC 8 Problems/Problem 25"
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Revision as of 16:18, 25 January 2024
Problem
A small airplane has rows of seats with
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 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, assume 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 grand total of cases. The final probability is
.
~blueprimes [1]
Solution 2 (Straightforward Casework)
Suppose the passengers are indistinguishable.
What this question is asking, is really, if 4 empty seats are places, 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
~rhydon516