**(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 | 2 years agoHow many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

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**(i)** 4 letters are used at a time

**(ii)** all letters are used at a time

**(iii)** all letters are used but first letter is a vowel ?

There are 10 persons named P_{1}, P_{2}, P_{3} …, P_{10}. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement P_{1} must occur whereas P_{4} and P_{5} do not occur. Find the number of such possible arrangements.

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