Here, even digits are 2 and 4.

Number of digit "2" presents = 2

Number of digit "4" presents = 2

$$\therefore$$ Total even digits = 4

$$\therefore$$ Total 4 even places presents.

Number of ways to put those 4 digits in those 4 places $$ = {{{}^4{C_4} \times 4!} \over {2!2!}} = {{4!} \over {2!2!}}$$

Now, remaining 5 digits (three 1 and two 3) can be put in those 5 places in $$ = {{{}^5{C_5} \times 5!} \over {3!2!}} = {{5!} \over {3!2!}}$$ ways.

$$\therefore$$ Total possible 9 digit numbers

$$ = {{4!} \over {2!2!}} \times {{5!} \over {3!2!}} = 60$$