Number of possible eight digit number divisible by 9 An eight digit number divisible by 9 is o be formed by using 8 digits out of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition. Find the number of ways in which it can be done.
I know divisible rule of 9 is sum of all the digits should be multiple of nine. But don't know how to use it in permutation. 

The sum of all the numbers between $0$ and $9$ is $45$, hence a multiple of $9$.
This means that if you want the sum of $8$ numbers taken in this list to be a multiple of $9$, the sum of the two remaining numbers will also be a multiple of $9$.
The only possible couples are $(0,9),(1,8),(2,7),(3,6),(4,5)$.
When you select a given couple, you have then $8! = 40320$ possibilities to form your number with your $8$ digits.
Since you have $5$ different couples, the answer is $5*8! = 201600$.
If you forbid leading $0$, the number of possibilities become $7*7! = 35280$ for the last 4 couples. The answer then becomes $4*7*7+8! = 181440$.
