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Question: n cadets have to stand in a row. If all possible permutations are equally likely, then the probabili...

n cadets have to stand in a row. If all possible permutations are equally likely, then the probability that two particular cadets stand side by side, is

A

2n\frac { 2 } { n }

B

1n\frac { 1 } { n }

C

2(n−1)!\frac { 2 } { ( n - 1 ) ! }

D

None of these

Answer

2n\frac { 2 } { n }

Explanation

Solution

Total number of ways =n!= n !.

Favourable cases =2(n−1)!= 2 ( n - 1 ) !

Hence required probability =2(n−1)!n!=2n= \frac { 2 ( n - 1 ) ! } { n ! } = \frac { 2 } { n }.