In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?

Explanation

(i) There are remaining 10 letters in the words starting with P and ending with S in which T occurs twice.
No. of such words `= (10!)/(2!) = 1814400`.
(ii) If all vowels A,E,I,O,U are together then taking them as on letter, the number of ways of arranging, 8 letters in which T occurs twice
`=(8!)/(2!) = (40320)/(2) = 20160`
Again, no. of ways of arranging 5 vowels
`.^(5)P_(5) = 5! =120`
Therefore, total words in which vowels are together `= 20160 xx 120 = 2419200`
(iii) The 12 letters of the word PERMUTATIONS can filled at 12 places in the following way:
1 2 3 4 5 6 7 8 9 10 11 12
If there are always four letters between P and S then we can place P at 1,2,3,4,5,6, or 7th place and as a result S can be placed at 6,7,8,9,10,11 or 12th place.
So, no of ways to place `P = 7`
No. of ways to place `S = 7`
Total ways `= 7 +7 14`
No. of ways to place remaining 10 letterss at remaining 10 places in which T occur twice.
`= (10!)/(2!) = (3628800)/(2) = 1814400`
Therefore, required number of words
`= 1814400 xx 14`
`= 25401600`