Words with or without meaning are to be formed using all the...

We have,

$EXAMINATION$

$AAEIIMNNOTX$

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

Total number of ways to form words using the alphabets of $EXAMINATION$ is

$= \dfrac{11 !}{2 ! \cdot 2 ! \cdot 2 !}$

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

.$- - - M - - - -$

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

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