Question

$n$ 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. $\frac{2}{\mathrm{n}}$
• B. $\frac{1}{\mathrm{n}}$
• C. $\frac{2}{(\mathrm{n}-1)!}$
• D. None of these.

Answer 1

The correct option is A

$\frac{2}{\mathrm{n}}$

Explanation for the correct option:

Finding the probability:

We can arrange $n$ cadets in $n!$ ways.

Total cases $=n!$

Two particular cadets stand side by side $=\left(n-1\right)!$$\times 2!$

Favourable cases $$\begin{gathered} =2!(\mathrm{n}-1)! \\ =2(\mathrm{n}-1)! \end{gathered}$$

Hence, the required probability $=\frac{\text{Favourable cases}}{\text{Total no. of ways}}$

$=\frac{2(\mathrm{n}-1)!}{\mathrm{n}!}$

$=\frac{2}{\mathrm{n}}$

Therefore, option(A) is correct answer.