In how many ways can the letters of the word PERMUTATIONS be arranged if:

(i) words start with P and end with S, 
(ii) vowels are all together, 
(iii) there are always 4 letters between P and S?

In the word PERMUTATIONS, there are 12 letters in total, with 2 T’s and all other letters appearing only once.

(i) 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 to arrange.
Required number of arrangements = 10! / 2! = 1814400

(ii) There are 5 vowels (E, U, A, I, O), each appearing only once. Since they have to occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects. These 8 objects in which there are 2 T’s can be arranged in 8! / 2! ways. Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways.
Required number of arrangements = (8! / 2!) × 5! = 20160 × 120 = 2419200

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S. Therefore, the places of P and S are fixed. The remaining 10 letters in which there are 2 T’s can be arranged in 10! / 2! ways. Also, P and S can be placed such that there are 4 letters between them in 2 × 7 = 14 ways.
Required number of arrangements = (10! / 2!) × 14 = 1814400 × 14 = 25401600

Hence,
(i) 1814400 arrangements
(ii) 2419200 arrangements
(iii) 25401600 arrangements