Question

How many different nine-digit numbers can be formed from the number $223355888$ by rearranging its digits so that the odd digits occupy even positions?

Answer 1

Number of even places $=4$

Number of even digits $=5(2,2,8,8,8)$

Number of odd places $=5$

Number of odd digits $=4(3,3,5,5)$

Odd digits can arranged in $\frac{4!}{2!2!}$ ways $=6$ ways.

Even digits can be arranged in $\frac{5!}{2!3!}=10ways$ .

Hence the required number of ways $=6\times 10=60$ ways