In how many ways can the letters of the word PERMUTATIONS be arranged if the

(ii) Vowels are all together,

(iii) There are always 4 letters between P and S?

Asked by Pragya Singh | 1 year ago |  58

##### Solution :-

(i) Total number of letters in PERMUTATIONS =12

Only repeated letter is T; 2times

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

Number of letters remaining =12 – 2 = 10

Number of permutations =

$$\dfrac{^{10}_{10}P}{2!}=\dfrac{10!}{2(10-10)!}=\dfrac{10!}{2}$$

= 1814400

(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)

Now, we consider all the vowels together as one.

Number of permutations of vowels = 120

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

Number of permutations =

$$\dfrac{^{8}_{8}P}{2!}=\dfrac{8!}{2(8-8)!}=\dfrac{8!}{2}$$

= 20160

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 =

$$\dfrac{^{10}_{10}P}{2!}=\dfrac{10!}{2(10-10)!}=\dfrac{10!}{2}$$

= 1814400

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

Answered by Abhisek | 1 year ago

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