A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back
row of 2 seats . 12 passengers board the aeroplane.
(i) How many possible seating arrangements are there for the 12 passengers? Give your answer
correct to 3 significant figures. [2]
These 12 passengers consist of 2 married couples (Mr and Mrs Lin and Mr and Mrs Brown), 5 students
and 3 business people.
(ii) The 3 business people sit in the front row. The 5 students each sit at a window seat. Mr and Mrs
Lin sit in the same row on the same side of the aisle. Mr and Mrs Brown sit in another row on
the same side of the aisle. How many possible seating arrangements are there? [4]
(iii) If, instead, the 12 passengers are seated randomly, find the probability that Mrs Lin sits directly
behind a student and Mrs Brown sits in the front row.
Please explain
your answer. [4]
(i) There are 12! (12 factorial) possible seating arrangements for the 12 passengers. This is equal to 479,001,600.
(ii) There are 4! (4 factorial) possible seating arrangements for the 4 rows. This is equal to 24. Since the 5 students each sit at a window seat, there are 5! (5 factorial) possible seating arrangements for the 5 students. This is equal to 120. Since Mr and Mrs Lin sit in the same row on the same side of the aisle, there are 2! (2 factorial) possible seating arrangements for them. This is equal to 2. Since Mr and Mrs Brown sit in another row on the same side of the aisle, there are 2! (2 factorial) possible seating arrangements for them. This is equal to 2. Therefore, there are 24 x 120 x 2 x 2 = 69,120 possible seating arrangements.
(iii) The probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row is 1/69,120. This is because there is only one possible seating arrangement that satisfies this condition out of the 69,120 possible seating arrangements.
(i) To find the number of possible seating arrangements for the 12 passengers, let's break it down into steps:
Step 1: Arrange the 3 business people in the front row.
Since there are 3 people and 3 seats in the front row, there is only 1 possible arrangement for them.
Step 2: Arrange the 5 students in the window seats.
Since there are 5 students and 5 window seats available, they can be arranged in 5! = 120 different ways.
Step 3: Arrange Mr and Mrs Lin in the same row on the same side of the aisle.
Since they need to sit in the same row together, we can treat Mr and Mrs Lin as one entity. This entity can be placed in one of the 4 rows. And once it is placed, there are 2 seats available in that row to choose from. Therefore, there are 4 x 2 = 8 different ways to arrange Mr and Mrs Lin.
Step 4: Arrange Mr and Mrs Brown in another row on the same side of the aisle.
Similarly, treat Mr and Mrs Brown as one entity. This entity can be placed in one of the remaining 3 rows (excluding the row where Mr and Mrs Lin are seated). And once it is placed, there are 2 seats available in that row to choose from. Therefore, there are 3 x 2 = 6 different ways to arrange Mr and Mrs Brown.
Step 5: Arrange the remaining 2 passengers in the back row.
Since there are 2 passengers and 2 seats in the back row, there is only 1 possible arrangement for them.
To find the total number of possible seating arrangements, we multiply the number of possibilities from each step together:
Total possible seating arrangements = 1 x 120 x 8 x 6 x 1 = 5,760
(ii) In this case, we have some given conditions for the seating arrangements, which narrow down the possibilities.
The 3 business people sit in the front row, so there's only one possible arrangement for them.
The 5 students each sit at a window seat. Since there are 5 window seats, there's only one way to arrange the students in those seats.
Mr and Mrs Lin sit in the same row on the same side of the aisle. This means they can choose one row together, and there are 2 seats available in that row for them to choose from.
Mr and Mrs Brown sit in another row on the same side of the aisle. This means they can choose one row together, and there are 2 seats available in that row for them to choose from.
The remaining 2 passengers can sit in the back row. Since there are 2 seats in the back row, there's only one possible arrangement for them.
To find the total number of possible seating arrangements, we multiply the number of possibilities from each step together:
Total possible seating arrangements = 1 x 1 x 2 x 2 x 1 = 4
(iii) If the 12 passengers are seated randomly, we need to find the probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row.
Let's find the total number of possible seating arrangements first:
Total possible seating arrangements = 14! = 87,178,291,200 (using the formula for permutations)
Now, let's find the number of favorable outcomes:
Mrs Lin can choose any of the 12 remaining seats (excluding the student's seat) as her seat.
The chance that Mrs Lin chooses the correct seat directly behind a student is 1 out of 12.
Mrs Brown can choose any of the remaining 11 seats (excluding the student's seat and the front row) as her seat.
The chance that Mrs Brown chooses the front row is 1 out of 11.
Since the seating arrangements of the other passengers don't matter for this specific condition, we consider these as one possible arrangement.
Number of favorable outcomes = 12 x 1 x 11 x 1 = 132
Now, let's calculate the probability:
Probability = Number of favorable outcomes / Total possible seating arrangements
Probability = 132 / 87,178,291,200 ≈ 1.5 x 10^-12 (to 3 significant figures)
To answer this question, we will break it down into different parts and explain the steps to calculate each part.
(i) In the first part, we need to find the number of possible seating arrangements for the 12 passengers on the airplane. We know that there are 14 seats in total, and they are arranged in 4 rows of 3 seats and a back row of 2 seats.
To calculate the number of possible seating arrangements, we need to use the concept of permutations. We will calculate the permutations of 12 passengers taken from 14 seats.
The formula to calculate permutations is:
nPr = n! / (n - r)!
Where n is the total number of items and r is the number of items being chosen.
In this case, we have 14 seats (n) and we need to choose 12 passengers (r).
Calculating this using the formula gives us:
14P12 = 14! / (14 - 12)!
= 14! / 2!
= (14 * 13 * 12!) / 2!
= 2184
So, there are 2184 possible seating arrangements for the 12 passengers on the airplane.
(ii) In the second part, we need to consider the given restrictions on the seating arrangement. The 3 business people sit in the front row, the 5 students each sit at a window seat, Mr and Mrs Lin sit in the same row on the same side of the aisle, and Mr and Mrs Brown sit in another row on the same side of the aisle.
Let's break this down step by step:
1. The 3 business people sit in the front row: Since there is only one front row, the arrangement for the 3 business people is fixed and cannot be rearranged. So, we don't need to consider them when calculating the arrangements.
2. The 5 students each sit at a window seat: Since there are 5 window seats, we have 5 choices for the first student, 4 choices for the second student, and so on. So, the number of possible arrangements for the students is 5!
3. Mr and Mrs Lin sit in the same row on the same side of the aisle: Since they need to sit together, we can treat them as a single unit. We can place this unit on any of the remaining available seats. So, we have 9 remaining seats after considering the front row and the window seats. The number of arrangements for Mr and Mrs Lin is then 9!
4. Mr and Mrs Brown sit in another row on the same side of the aisle: Similar to Mr and Mrs Lin, we treat them as a single unit and place them on any of the remaining seats. So, the number of arrangements for Mr and Mrs Brown is also 9!
To calculate the total number of possible seating arrangements, we need to multiply the number of arrangements for each group: 5! * 9! * 9!
Calculating this gives us:
5! * 9! * 9! ≈ 1.366 × 10^9 (rounded to 3 significant figures)
So, there are approximately 1.366 × 10^9 possible seating arrangements that meet the given restrictions.
(iii) In the third part, we are asked to find the probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row.
To calculate this probability, we need to determine two things:
1. The total number of possible seating arrangements when the 12 passengers are seated randomly, which we calculated in part (i) to be 2184.
2. The number of seating arrangements where Mrs Lin sits directly behind a student and Mrs Brown sits in the front row.
To calculate the number of seating arrangements meeting these conditions, we treat Mrs Lin and the student directly in front of her as a single unit. This unit can be placed in any of the 12 remaining seats (excluding the front row). The rest of the passengers can then be arranged in the remaining 11 seats.
So, the number of seating arrangements meeting the conditions is 12! * 11!
The probability is then given by:
Probability = Number of seating arrangements meeting the conditions / Total number of possible seating arrangements
Probability = (12! * 11!) / 2184
Calculating this gives us the probability.