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The number of permutation of all the let...

The number of permutation of all the letters of the word `PERMUTATION` such that any two consecutive letters in the arrangement are neither both vowels nor both identical is

Text Solution

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The correct Answer is:
`57xx(5!)^(2)`
Number of arrangements with no two vowels together
=(Arranging P,R,M,T,T,N)
`xx` (Selecting 5 gaps from 7 created for vowels )
`xx` (Arranging E, U, A, I, O in five gaps)
`=(6!)/(2!)xx .^(7)C_(5)xx5!`
Number of arrangements which contains 2T's together
=(Arranging P,R, M,(T,T)N)
`xx` (Selecting 5 gaps from 6 created for vowels)
`xx`(Arranging E, U, A,I,O in five gaps)
`=5!xx .^(6)C_(5)xx5!`
`therefore` Required arrangements `=(6!)/(2!)xx .^(7)C_(5)xx5!-5!xx .^(6)C_(5)xx5!`
`=57xx(5!)^(2)`
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