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  4. N cadets have to stand in a row if all possible permutations are equally likely, the probability of two particular cadets standing side by side is:

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Q. N cadets have to stand in a row if all possible permutations are equally likely, the probability of two particular cadets standing side by side is:

Permutations and Combinations

Solution:

We can arrange N persons in a row in N! ways.
We assume two particular cadets as one.
$\therefore $ Remaining (N - 1) persons can be arranged in (N - 1)! ways.
Also, these 2 cadets can be arranged themselves in 2! ways.
Therefore, P (2 particular cadets standing side by side)
$ = \frac{(N - 1)! \times 2!}{N!} = \frac{2}{N}$