An eight digit number divisible by 9 is to  be formed by using 8 digits out of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without replacement. The number of ways in which this can be done is

We have 0 + 1 + 2 + 3 … + 8 + 9 = 45

To obtain an eight digit number exactly divisible by 9, we must not use either (0, 9) or (1, 8) or (2, 7) or (3, 6) or (4, 5). [Sum of the remaining eight digits is 36 which is  exactly divisible by 9.]

When, we do not use (0, 9), then the number of required 8 digit numbers is $8!$.

When one of (1, 8) or (2, 7) or (3, 6) or (4, 5) is not used, the remaining digits can be arranged in $8!-7!$ ways.{0 can not be at extreme left.}

Hence, there are $8!+4(8!-7!)=\left(36\right)(7!)$ numbers in the  desired category.