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 $2T'\,s$ 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
$=\frac{6 !}{2 !} \times{ }^{7} C_{5} \times 5 !-5 ! \times{ }^{6} C_{5} \times 5 !$
$=57 \times(5 !)^{2}$