Question

In how many ways can the letters of the word $DIRECTOR$ be arranged so that the three vowels are never together?

Answer 1

WORD$\to$ DIRECTOR

Total no. of arrangements

$\frac{8!}{2!}$

Total no. of arrangements when $3$ vowels are together

$=\frac{6!}{2!}\times 3!$

Required ways$=\frac{8!}{2!}-\frac{6!}{2!}\times 3!$

$=\frac{6!}{2!}[56-6]$

$=\frac{50\times 6!}{2}$

$=18000$.