Question

The number of 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

Answer: (B)

Number of arrangements with no two vowels together
= ( Arranging $P,R,M,T,T,N$ )
$\times ($ Selecting $5$ gaps from $7$ created for vowels)
$\times ($ Arranging $E,U,A,I,O$ in five gaps $)$
$=\frac{6!}{2!}\times {}^7{C}_5\times 5!$
In above number of arrangements when $2T^{\prime }s$ come together
=( Arranging $P,R,M,(T,T),N$ )
$\times ($ Selecting $5$ gaps from $6$ created for vowels)
$\times$ ( Arranging $E,U,A,I,O$ in five gaps )
$=5!\times {}^6{C}_5\times 5!$
$\therefore$ Required arrangements
$=\frac{6!}{2!}\times {}^7{C}_5\times 5!-5!\times {}^6{C}_5\times 5!$
$=57\times (5!{)}^2$