Difference between revisions of "2024 AMC 8 Problems/Problem 25"
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<math>\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}</math> | <math>\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}</math> | ||
+ | ==Video Solution (4 minutes) by MegaMath== | ||
+ | https://www.youtube.com/watch?v=DgQsljPaE5Y | ||
==Solution 1 (Complementary Counting Casework)== | ==Solution 1 (Complementary Counting Casework)== |
Revision as of 11:53, 27 January 2024
Contents
- 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 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?
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 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 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
~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 total combinations for this case.
Case 2: Three people sitting in middle seats.
In this case, there are 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
total combinations.
Case 3: Two people sitting in middle seats.
In this case, there are 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
total combinations.
Case 4: One person sitting in a middle seat
In this case, there are 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
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
combinations, or 195 combinations. The final step is to find the total amount of combinations without restrictions. This is simply . So, finally employing complementary counting, we have that the probability that there will be 2 adjacent seats for the couple is
~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
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.