Question

Suppose that n people are seated in a random manner in a row of n theater seats. What is the probability that two particular people A and B will be seated next to each other?

Can you provide a detailed explanation on the steps and process you use? Do you use combinations or permutations to solve for the answer? The answer should be n/2, according to my book, but I don't know how to get it.

Answer 1

This is a case of permutations, which states that the arrangements of n distinct things, all taken together = n!

We know that Probability = Favourable Outcomes/Total Outcomes

Total outcomes: Putting n people in n seats in a row = n!.

Favourable outcomes: We want 2 people A and B together. When such cases come, when you need 2 people or 3 people together or in general x out of n people together, we consider x peopleas 1 person. Then use the factorial rule the number of arrangements = (n - x + 1)! * x!

Eg A B C D and E to be seated in a row, where A and B are always together Then you have AB C D E = 4 people, who can be seated in 4! ways and A and B can interchange in 2! = 2 ways = AB or BA.

Therefore no of ways = 4! * 2!

we see n = 5, x = 2, therefore (n - x + 1)! * x! = (5 - 2 + 1)! * 2! = 4! * 2!

Therefore when seating n people in a row, such that A and B need to be together, the total favourable outcomes = (n - 2 + 1)! * 2! = (n-1)!2!

Therefore the required probability = (n-1)! * 2!/n!

n! = n * (n - 1)! (eg 6! = 6 * 5!) and 2! = 2

Therefore (n-1)! * 2/[n * (n-1)!]

= 2/n