Question

The total number of nine-digit numbers that can be formed using the digits $2,2,3,3,5,5,8,8$ and $8$ so that the odd digit occupy the even places is

Answer 1

Number of ways in which odd digits $3,3,5$ and $5$ can occupy four even places is equal to $\frac{4!}{2!2!}=6$

Number of ways in which rest five digits $2,2,8,8$ and $8$ can occupy rest of the five places is equal to $\frac{5!}{2!3!}=10$

Hence, the required numbers of ways is $6\times 10=60$