Question

In how many ways can the letters of the word 'PERMUTATIONS' be arranged if the
(i) words start with $P$ and ends with $S$ and

(ii) Vowels are all together.

Answer 1

$\left(i\right)$ Words start with $P$ and end with $S:$

When we fix $P$ at the left end and $S$ at the right end, then we are left with $10$ letters.

Hence, required no. of ways $=\frac{10!}{2!}=1814400$

$\left(ii\right)$ Vowels are all together:

There are $5$ vowels in the given word, $1E,1U,1A,1I$, and $1O$.

Since these vowels are occurring together, so consider them as one letter, and when this letter is combined with the remaining $7$ letters, then we have $8$ letters in all, which can be arranged in $\frac{8!}{2!}$ ways.

Corresponding to the arrangements, the $5$ vowels can be arranged in $5!$ ways.

Hence, required no. of ways $=\frac{8!}{2!}\times 5!=20160\times 5!=24,19,200$