Question

The number of different ordered permutations 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

• A. $63\times 6!\times 5!$
• B. $57\times 5!\times 5!$
• C. $33\times 6!\times 5!$
• D. $7\times 7!\times 5!$

Answer 1

The correct option is B $57\times 5!\times 5!$

PERMUTATION
There are 5 vowels and 6 consonants (2 T's)
Possible permutations of PRMTTN
$=\frac{6!}{2!}$
There are 7 gaps between 6 consonants and no. of ways in which they can be occupied by 5 vowels
$={}^7{C}_5\times 5!$

Total ways $=\frac{6!}{2!}\times {}^7{C}_5\times 5!$

$=3\times 5!\times 21\times 5!$

$=63\times 5!\times 5!$

But these ways also contain words in which $2$ T's are together

Now assume TT to be one letter

$\therefore$ No. of ways $=$ No. of ways of PRMTTN $\times$ No. of ways vowels can be placed

$=5!\times {}^6{C}_5\times 5!$$=6\times 5!\times 5!$

$\therefore$ Required no. of ways $=63\times 5!\times 5!-6\times 5!\times 5!$

$=57\times 5!\times 5!$