JEE Main 2021 — Probability Question with Solution

We have,

$\mathrm{EXAMINATION}$

$\mathrm{AAEIIMNNOTX}$

Given word has $A\text{'}s\to 2, I\text{'}s\to 2, N\text{'}s\to 2$ and rest other are distinct alphabets.

Total number of ways to form words using the alphabets of $\mathrm{EXAMINATION}$ is

$=\frac{11!}{2!·2!·2!}$

Now, putting $M$ at $4^{\mathrm{th}}$ position

.$- - - M- - - -$

Total words with $M$ at fourth Place $=\frac{10!}{2! 2! 2!}$

Required probability $=\frac{10!}{11!}=\frac{1}{11}$