n cadets have to stand in a row. If all possi | vedclass

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(A) The total number of ways to arrange $n$ cadets in a row is $n!$.To find the number of favorable cases where two particular cadets stand side by si

Vedclass · Accepted Answer

$\frac{2}{n}$

Vedclass · Answer

(A) The total number of ways to arrange $n$ cadets in a row is $n!$.To find the number of favorable cases where two particular cadets stand side by side,we treat these two cadets as a single unit.Now,we have $(n - 1)$ units to arrange,which can be done in $(n - 1)!$ ways.Within the single unit,the two particular cadets can be arranged in $2! = 2$ ways.Thus,the total number of favorable arrangements is $2 	imes (n - 1)!$.The required probability is the ratio of favorable cases to the total number of cases:$P = \frac{2 	imes (n - 1)!}{n!} = \frac{2 	imes (n - 1)!}{n 	imes (n - 1)!} = \frac{2}{n}$.

$n$ 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

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