**Solutions:**

**(i) 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 = (10!)/2! = **1814400**

**(ii) Vowels are all together:**

There are 5 vowels in the given word, 1 E, 1 U, 1 A, 1 I, and 1 O. 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 (8!)/(2!) ways. Corresponding to the arrangements, the 5 vowels can be arranged in 5! ways

Hence, required no. of ways = (8!)/(2!) * 5! = 20160 * 5! = **2419200**

**(iii)** There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice.

Now first we need to see how many ways we can make word with 4 letter between P and S.

Except P and S there are total of 10 letters, so number of way of selecting them = 10C4 = 210

Also note that question is asking to place exactly 4 words between P and S, but does not tells you if P has to be the first letter of S has to be the first letter. So In all the above combinations, we can rotate the position of P and S.

So total way = 210*2 = 420

The selected 4 letters can be rotated between P and S in = 4! ways

So total ways = 420 * 4!

Consider this 6 letter chunk (P, S, and 4 letter between them) as 1 letter.

Remaining letters are 6. So in total we have 7 letters, which can be arranged in 7! ways.

So total number of ways = 7! * 420 * 4!

Now since letter T was repeated twice, we should divide the above result by 2!.

So Total number of ways = 7! * 420 * 4! / 2! = 25401600

**another approach for (ii)**

**There are always 4 letters between P and S:**

Here P and S are fixed, so 10 letters are left, out of which 2 are T's

So, 10 letters out of which 2 are T's can be arranged in 10!/2! = 1814400 ways

Now letters P and S can be arranged so that there are 4 letters between them, which can be done in 2*7 = 14 ways

Required no. of ways = 1814400 * 14 = **25401600**