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.
On
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 $4*7*7+8! = 181440$.
On
Eight digit no divisible by $9$ i.e. sum of digits divisible by $9$
$(1.)$ Total no formed by $(1,2,3,4,5,6,7,8) = 8!$
$(2.)$ Total no formed by $(0,2,3,4,5,6,7,9) = 7\times7!$
$(3.)$ Total no formed by $(1,0,3,4,5,6,9,8) = 7×7!$
$(4.)$ Total no formed by $(1,2,0,4,5,9,7,8) = 7×7!$
$(5.)$ Total no formed by $(1,2,3,0,5,6,7,8) = 7×7!\, 8! + 28 × 7 ! = 36 × 7 !$
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 $8!+(4)(7)(7!)$.