For those three odd digit numbers 1, 1, 3 we can choose any three positions out of the four even positions.

$$ \therefore $$ No of ways we can choose 3 positions out of the 4 positions = ⁴C₃

After choosing those three positions, number of ways we can arrange three odd digit numbers = ⁴C₃ $$ \times $$ $${{3!} \over {2!}}$$

Then the remaining 6 digits can be arrange = $${{6!} \over {2!4!}}$$ ways

$$ \therefore $$ Total number of 9 digit numbers = ⁴C₃ $$ \times $$ $${{3!} \over {2!}}$$ $$ \times $$ $${{6!} \over {2!4!}}$$ = 180