Class 11

Math

Algebra

Permutations and Combinations

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

- 25401600

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 $2!_{10}C_{4} $ 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 $=2!2!_{10}C_{4} 4!7!$

$=25401600$