Total number of 6 digit number in which only and all the five digits 1,3,5,7 and 9 appear is

Given 5 digits $=1,3,5,7,9$

For selection of one digit we have ${ }^5{C}_1$ choice.

And six digits can be arranged in $\frac{6!}{2!}$ ways

$\text{ Total }=5\times \frac{6!}{2!}$