Question

There are always $4$ letters between $P$ and $S$?

Answer 1

There are $12$ letters in word $PERMUTATIONS$. $T$ occurs twice .
Here, $P$ and $S$ are fixed. These $P$ and $S$ can interchange their position. Hence, $P$ and $S$ can be arranged in $2!$ ways.
Since, $P$ and $S$ are fixed, there are $10$ letters left , where $T$ occurs twice .
From these $10$ letters, $4$ letters can be chosen in $\frac{{}^{10}{C}_4}{2!}$ ways.
Now since, $4$ letters are between $P$ and $S$, so these six letters can be considered as a single object (letter). These four letter (between $P$ and $S$ )can be arranged in $4!$ ways.
Now, the remaining $6$ letters and 1 object i.e. total $7$ can be arranged in $7!$ ways.
Hence, required number of ways $=2!\frac{{}^{10}{C}_4}{2!}4!7!$
$=25401600$