Official solution
Here none number repeats more than once.
We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in 5C1 ways.
Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.
We can arrange those six digits in \[{{6!} \over {2!}}\] ways.
\[ \therefore \] Total six digit numbers = 5C1 \[ \times \] \[{{6!} \over {2!}}\] = \[{5 \over 2}\left( {6!} \right)\]