Question

Seven persons sit in a row at random. The probability that three persons $A,B$ and $C$ sit in that order, not necessarily side by side, is

• A. $\frac{3}{7!}$
• B. $\frac{{}^7{C}_3}{7!}$
• C. $\frac{1}{3!}$
• D. $\frac{1}{7!}$

Answer 1

Answer: (C)

$\frac{1}{3!}$

Since our only concern is about $A,B$ and $C$, we look at their permutations.
The total number of ways they can sit irrespective of the other people is $3!$ number of ways.
Given the constraint, for the same case of the other $4$ people, $A,B$ and $C$, have only $1$ way to sit.
So, for every case of desired permutation, there are $3!$ times more permutations in the universal set.
So probability : $\frac{1}{3!}$