Question

There are N passengers in a plane with N assigned seats (N is a positive integer), but after boarding, the passengers take the seats randomly. Assuming all seating arrangements are equally likely, what is the probability that no passenger is in their assigned seat? Compute the probability when N → [infinity]

Answer 1

Answer:

The probability of no passenger  in his assigned seat is:

$\frac{(N-1)!}{N!} = \frac{1}{N}$

When N→∞, this probability is 0.

Step-by-step explanation:

Hi!

Each possible ways in which the passengers take seats can be represented as a permutation of the numbers between 1 and N. We can say that the "right" way (each passenger in its assigned seat) is: A = 12345....N. This means passenger 1 in seat 1, passenger 2 in seat 2, etc.

There N! permutations in total, and we have to count which ones have each number in a different position that in A. For example, in the case N=4:

2413 is one permutation in which no passenger is in his assigned seat

How many of those permutations exist? In the first seat, we can put any number but not 1, so there are (N-1) possibilities. Then in the second seat we can put (N-2) possible numbers, beacuse we cannot put nor 2 nor the number in the first place. If we continue thinking like this, we get (N-1)! permutations in which no passenger is in his assigned seat.

Then the probability of no passenger in his assigned seat is:

$\frac{(N-1)!}{N!} = \frac{1}{N}$

When N→∞, this probability is 0.