There are 12 letters of which T appears 2 times (i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times. ∴ The required words = $\frac{10!}{2!}$ = $\frac{10\times 9\times 8\times 7\times \times 5\times 4\times 3\times 2!}{2!}$ = 1814400 (ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there are 2Ts, which can be rearranged in $\frac{8!}{2!}$ = 20160 ways. Corresponding to each of these arrangements the 5 vowels E, U, A, I, O can be rearranged in 5! = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 × 120 = 2419200. (iii) When there are always 4 letters between P & S ∴ P & S can be at 1st & 6th place 2nd & 7th place 3rd & 8th place 4th & 9th place 5th & 10th place 6th & 11th place 7th & 12th place. So, P & S will be placed in 7 ways & can be arranged in 7 × 2! = 14 The remaining 10 letters with 2T’s, can be arranged in $\frac{10!}{2!}$ = 1814400 ways ∴ The required number of arrangements = 14 × 1814400= 25401600.