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:

Question

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: (A) (4 /N) (B) (3/N2) (C) (1/2N) (D) (2/N)

Tardigrade Admin · Accepted Answer

We can arrange N persons in a row in N! ways.
We assume two particular cadets as one.

∴

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}

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:

$\frac{4}{N}$

$\frac{3}{N ^{2}}$

$\frac{1}{2 N}$

$\frac{2}{N}$

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:

N4​

N23​

2N1​

N2​

We can arrange N persons in a row in N! ways. We assume two particular cadets as one. ∴ 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) =N!(N−1)!×2!​=N2​

$\frac{4}{N}$

$\frac{3}{N ^{2}}$

$\frac{1}{2 N}$

$\frac{2}{N}$