Difference between revisions of "2024 AMC 8 Problems/Problem 25"
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~NTfish | ~NTfish | ||
− | ==Solution 4 (Permutations)== | + | ==Solution 4 (Permutations, Fastest + Simplest Written Solution)== |
There are <math>12\cdot 11 = 132</math> for two people (the married couple) to be seated. This will be our denominator. | There are <math>12\cdot 11 = 132</math> for two people (the married couple) to be seated. This will be our denominator. | ||
Revision as of 09:59, 7 February 2024
Contents
[hide]- 1 Problem
- 2 Video Solution (Fastest) by MegaMath
- 3 Solution 1 (Complementary Counting Casework)
- 4 Solution 2 (Straightforward Casework)
- 5 Solution 3 (Complementary Casework on Middle Seats)
- 6 Solution 4 (Permutations, Fastest + Simplest Written Solution)
- 7 Video Solution 1 by Math-X (First understand the problem!!!)
- 8 Video Solution 2 by OmegaLearn.org
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by NiuniuMaths (Easy to understand!)
- 11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 12 Video Solution by Interstigation
- 13 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 (Fastest) 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
Solution 4 (Permutations, Fastest + Simplest Written Solution)
There are for two people (the married couple) to be seated. This will be our denominator.
There are pairs of seats that are next to each other in the diagram ( per row; left-middle and middle-right). This will be our numerator.
Since there are total people on the plane, we should multiply our numerator by that to account for all ways the 10 people could be seated (e.x. the husband and the wife could be switched around and it would still work, same applies to the other passengers)
Therefore, our numerator is .
This creates the fraction , which simplifies to
- Siddharth Mirchandani (svm2020), John Adams Middle School
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=mCibdgEbhXc7hd9t&t=8376 ~Math-X
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=ArN4qVlBDTM
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=PBNyuSg0GX4
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=bxZHXPMcsGI
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=3297
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.