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?

Total number of letters in PERMUTATIONS =12

Only repeated letter is T ; 2times

(i) First and last letter of the word are fixed as P and S respectively.

Number of letters remaining =12-2=10

⇒ No. of permutations =

(ii) No. of vowels in PERMUTATIONS = 5 (E,U,A,I,O)

Now, we consider all the vowels together as one.

Number of permutations of vowels =

Now total number of letters = 12-5+1=8

⇒ No. of permutations =

Therefore, total number of permutations = 120× 20160=2419200

(iii) Number of places are as 1 2 3 4 5 6 7 8 9 10 11 12

There should always be 4 letters between P and S.

Possible places of P and S are 1 and 6, 2and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12

Possible ways =7,

Also, P and S can be interchanged,

No. of permutations =2× 7 =14

Remaining 10 places can be filled with 10 remaining letters,

∴ No. of permutations =

Therefore, total number of permutations = 14 × 1814400 =25401600.