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

In the word PERMUTATIONS, there are 2 Ts and all the other letters appear 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.

Hence, in this case, required number of arrangements =10!/2!=1814400

(ii) There are 5 vowels in the given word, each appearing only once.

Since they have to always 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 Ts can be arranged in 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

8!/2! x 5!=2419200

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S.

Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 Ts can be arranged in 10!/2! ways.

Also, the letters P and S can be placed such that there are 4 letters between them in 2 × 7 = 14 ways.

Therefore, by multiplication principle, required number of arrangements in this case

=10!/2! x 14=25401600