Difference between revisions of "2024 AMC 8 Problems/Problem 25"
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A small airplane has <math>4</math> rows of seats with <math>3</math> 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? | A small airplane has <math>4</math> rows of seats with <math>3</math> 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== | + | ==Solution 1 (Casework)== |
| − | + | ||
| + | Suppose the passengers are indistinguishable. There are <math>\binom{12}{8} = 495</math> total ways to distribute the passengers. We proceed with complementary counting, and instead, will count the number of passenger arrangement such that the couple cannot sit anywhere. Consider the partitions of <math>8</math> among the rows of <math>3</math> seats, to make our lives easier, assume they are non-increasing. We have <math>(3, 3, 2, 0), (3, 3, 1, 1), (3, 2, 2, 1), (2, 2, 2, 2)</math>. | ||
| + | |||
| + | For the first partition, clearly the couple will always be able to sit in the row with <math>0</math> occupied seats, so we have <math>0</math> cases here. | ||
| + | |||
| + | For the second partition, there are <math>\frac{4!}{2!2!} = 6</math> ways to permute the partition. Now the rows with exactly <math>1</math> passenger must be in the middle, so this case generates <math>6</math> cases. | ||
| + | |||
| + | For the third partition, there are <math>\frac{4!}{2!} = 12</math> ways to permute the partition. For rows with <math>2</math> passengers, there are <math>\binom{3}{2} = 3</math> ways to arrange them in the row so that the couple cannot sit there. The row with <math>1</math> passenger must be in the middle. We obtain <math>12 \cdot 3^2 = 108</math> cases. | ||
| + | |||
| + | For the fourth partition, there is <math>1</math> way to permute the partition. As said before, rows with <math>2</math> passengers can be arranged in <math>3</math> ways, so we obtain <math>3^4 = 81</math> cases. | ||
| + | |||
| + | Collectively, we obtain a grand total of <math>6 + 108 + 81 = 195</math> cases. The final probability is <math>1 - \frac{195}{495} = \boxed{\textbf{(C)}~\frac{20}{33}}</math>. | ||
| + | |||
| + | ~blueprimes | ||
Revision as of 14:02, 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 (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 arrangement 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
