Suppose we have to find the number of ways of rearranging the letters of the word $ROOT$.
In this case, the letter of the word are not all different. There are 2 $O$s.
Let us take the 2 $O$s as different say $O_{1}$  and $O_{2}$ .
Number of permutations of 4-different letters, taken all at a time = 4!
Consider one of these permutations say, $ROOT$. Corresponding to this permutation, we have 2! permutations $RO_{1}O_{2}T$ and  $RO_{2}O_{1}T$.
Therefore, the required number of permutations =$\frac{4!}{2!}=\frac{4\times 3\times 2!}{2!}$ $=3\times 4=12$

Consider the number of ways of rearranging the letters of the word $INSTITUTE$.
There are 9 letters and can be arranged in 9! ways.
∴  the number of permutations of 9 different letters taken all at a time is 9!.
$I$ appears 2 times and can be arranged in 2! ways.
$T$ appears 3 times and can be arranged in 3! ways.
∴  (2!*3!) permutations will be just the same permutation corresponding to chosen word.
Hence, total number of different permutations will be $\frac{9!}{2!3!}$

The number of permutations of $n$ objects, where $p$ objects are of the same kind and rest are all different=$\large \frac{n!}{p!}$

The number of permutations of $n$ objects taken all together, when $p$ of the objects are one kind, $q$ are of second kind and $r$ are of third kind... and the rest, if any, are of different kind is $\large \frac{n!}{p!q!r!}$