Question

There are $n$ persons $(n\ge 3)$, among whom are $A$ and $B$, who are made to stand in a row in random order. Probability that there is exactly one person between $A$ and $B$ is

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

Answer 1

The correct option is B $\frac{2(n-2)}{n(n-1)}$

person that must stand between $A$ and $B$ can be chosen in $(n-2)$ ways.
Now number of ways in which $x$ persons can be made to stand so that there is exactly one person in between $A$ and $B$ is equal to $(n-2).2.(n-2)!$
Also, total number of ways in which persons can be made to stand $=n!$
$\therefore$ Required probability $=\frac{2(n-2)(n-2)!}{n!}=\frac{2(n-2)}{n(n-1)}$