How many different nine digit numbers can be formed from the number...

Here we have 4 odd digits $(3,3,5,5)$ and 5 even digits (2, 2, 8, 8, 8). $\bar{O} \bar{E} \bar{O} \bar{E} \bar{O} \bar{O} \bar{E} \bar{O}$ where, $E \rightarrow$ even place and $O \rightarrow$ odd place \Rightarrow Number of ways odd digits will be place on even places $=\frac{4!}{2!2!}$ Number of ways even digits will be place on odd places $$\begin{aligned} & \quad=\frac{5!}{2!3!} \\ & \therefore \text{ Total number of ways }=\frac{4!}{2!2!} \times \frac{5!}{2!3!}=60 \end{aligned}$$