Question

In how many ways can the letters of the word $PERMUTATIONS$ be arranged if the.i) Words start with $P$ and end with $S$.ii) vowels are all togetheriii) there are always $4$ letters between $P$ and $S$

Answer 1

i) There are total $12$ letters in the word $PERMUTATIONS$, with $T$ repeated twice.

If $P$ and $S$ are fixed at the extreme ends ($P$ at the left end and S at the right end), then $10$ letters are left.

These ten letters with $T$ occurring twice can be arranged in $\frac{10!}{2!}$.

Hence, required number of arrangements =$\frac{10!}{2!}=1814400$

ii) There are total $12$ letters in the word $PERMUTATIONS$, with $T$ repeated twice.

Number of vowels in the given word are $5$.

Since vowels have to always occur together, so they are considered as a single object .

This single object (letter) together with the remaining $7$ letters will give us $8$ objects (letters).

These $8$ objects in which there are $2$ $T^{\prime }s$ can be arranged in $\frac{8!}{2!}$ ways .Corresponding to each of these arrangements, the $5$ different vowels can be arranged in $5!$ ways.

Therefore, by multiplication principle, required number of arrangements in this case $=\frac{8!}{2!}\times 5!=2419200$

iii) 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 $=2!\frac{{}^{10}{C}_4}{2!}4!7!$

$=25401600$