In how many ways can the letters of the word 'PERMUTATIONS' be arranged, if
(i) the words start with P and end with S?
(ii) the vowels are all together?

Explanation

(i) Let us fix P at the left end and S at the right end of each arrangement.
Then, we are left with 10 letters, out of which T occurs 2 times and the rest are all different.
Hence, the required number of arrangements `=(10!)/(2!)=1814400.`
(ii) The given word contains 5 vowels, namely E, U, A, I, O.
Treating these five vowels EUAIO as one letter, we have to arrange 8 letters, namely EUAIO + PRMTTNS, out of which T occurs 2 times and the rest are all different.
Number of all such arrangements `=(8!)/(2!)=20160.`
Now, the 5 vowels are all different. They can be arranged among themselves in 5! = 120 ways.
Hence, the number of arrangements in which 5 vowels are together`=(20160xx120)=2419200.`