A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back

Question

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

Answer 1

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.