Mathematics - Permutation Combination Question with Solution | TestHub

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, $E\to$ even place and $O\to$ 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 $\begin{array}{rl}& =\frac{5!}{2!3!}\\ & \therefore \text{ Total number of ways }=\frac{4!}{2!2!}\times \frac{5!}{2!3!}=60\end{array}$