QJEEM19Mae77d: JEE Main 2019 Mathematics - Class 11 JEEM19Mae77d Solution: All possible numbers are formed using the digits 1, 1, 2, 2, 2, …

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