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

\frac{2}{n}

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\frac{1}{n}

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\frac{2}{(n - 1)!}

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N o n e o f t h e s e

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Solution

The correct option is A $\frac{2}{n}$
Total number of ways = n!
Favourable cases = 2(n-1)!
Hence required probability $= \frac{2 (n - 1)!}{n!} = \frac{2}{n}$ .

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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

The correct option is A 2nTotal number of ways = n! Favourable cases = 2(n-1)! Hence required probability =2(n−1)!n!=2n.

The correct option is A $\frac{2}{n}$
Total number of ways = n!
Favourable cases = 2(n-1)!
Hence required probability $= \frac{2 (n - 1)!}{n!} = \frac{2}{n}$ .