Question

How many different nine digit number can be formed from the number $2,2,3,3,5,5,8,8,8$ by rearranging its digits so that the odd digits occupy even positions?

Answer 1

The four digits $3,3,5,5,$ can be arranged at four even places in $\frac{4!}{2!2!}=6$ ways

and the remaining digits $2,2,8,8,8$ can be arranged at the five odd places in $\frac{5!}{2!3!}=10$ ways.

Thus, the number of possible arrangements is $\left(6\right)\left(10\right)=60$.