Difference between revisions of "2024 AMC 8 Problems/Problem 25"
Blueprimes (talk | contribs) |
Blueprimes (talk | contribs) (→Solution 1 (Casework)) |
||
Line 7: | Line 7: | ||
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 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 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 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. | 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>. | 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 [https://artofproblemsolving.com/community/user/1096417] | ~blueprimes [https://artofproblemsolving.com/community/user/1096417] |
Revision as of 15:09, 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 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]