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

Options

A. `63xx6!xx5!`
B. `57xx5!xx5!`
C. `33xx6!xx5!`
D. `7xx7!xx5!`

Answer

B

Explanation

`(b)` The letters other than vowels are : `PRMTTN`
Number of permutations with no two vowels together is
`(6!)/(2!)xx^(7)C_(5)xx5!`
Further among these permutations the number of cases in which `T'`s are together is `5!xx^(6)C_(5)xx5!`
So the required number
`=(6!)/(2!)^(7)C_(5)xx5!-5!xx^(6)C_(5)xx5!=57xx(5!)^(2)`