Question

The number of possible arrangements of letters of the word $REVISION$ such that there are exactly two vowels between $E$ and $V$ is

• A. $4032$
• B. $480$
• C. $600$
• D. $720$

Answer 1

The correct option is A $4032$

WORD $=REVISION$

No.of vowels = $4=\{E,I,I,O\}$

No.of Consonants = $4=\{R,V,S,N\}$

${}^8{C}_2$ for selecting two positions for $E,V$

$2!$ for arranging them.

$4!$ for arranging rest letters.

two cases $(I,O)(I,I)$

Total no. of ways $={}^8{C}_2\times 2!\times 4!{+}^8{C}_2\times 2!\times 2!\times 4!$

$=4032$ Ans