Question

How many different words can be formed with the letters of the word $ORDINARY$ such that the vowels occupy odd places?

• A. $2880$
• B. $2400$
• C. $1500$
• D. $1440$

Answer 1

The correct option is D $1440$

There are $4$ place available for $3$ vowels so total possible arrangement$=4\times 3!=24$
Total number of arrangement for consonent =$\frac{5!}{2!}=60$
So,total number of words=$60\times 24=1440$