Question

The number of permutations that can be made out of the letters of the word “EQUATION” which start with a consonant and end with a consonant is

• A. $2!6!$
• B. $3!6!$
• C. $3!5!$
• D. $2!5!$

Answer 1

The correct option is B $3!6!$

Consonants occupy 2 ends in ${}^3{\text{P}}_2$ ways remaining 6 letters occupy 6 places in 6! Ways
So the required number of arrangements = ${}^3{\text{P}}_2.6!=3!6!$