Number of possible eight digit number divisible by 9

As pointed out in the post, all that matters is the digit sum. The sum of all your digits is divisible by $9$. So our number is divisible by $9$ if and only if the two digits not chosen are $0,9$ or $1,8$, or $2,7$, or $3,6$, or $4,5$.

If the two digits not chosen are $0,9$, there are $8!$ possible numbers.

If the two digits not chosen are any of the $4$ other pairs, then $0$ was chosen. Then there are in each case $7\cdot7!$ numbers. For $0$ cannot be the first digit of an $8$-digit number.

This gives a total of $8!+(4)(7)(7!)$.