How many phone number with $8$ digits exist s.t divide $2,3,5$ and there is no repetitive digit in it? Here is my approach:  
The last digit should be $0$ and the first digit does not $0$. Hence there are $9$ choices for the first digit, $8$ for second,...,$3$ for seventh.
So there are $9.8.7.6.5.4.3$ phone numbers that they don't divide $2,5$.  
Now I divide $9.8.7.6.5.4.3$ by $3$ and get approximation $9.8.7.6.5.4$
Because from each of three number (statistically!) one them is divisible by $3$.  
But what is the exact answer?

Recall that a number is divisible by $3$ if and only if its digit sum is divisible by $3$. We will be choosing $7$ digits from $1,2,\dots,9$ whose digit sum is divisible by $3$. 
The sum of the digits $1$ to $9$ is divisible by $3$, so the two digits we do not choose must have sum divisible by $3$. There are $12$ ways to do this, and then $7!$ ways to arrange them in a row, for a total of $12\cdot 7!$.
To see there are $12$ ways to decide which two digits we do not use, note that we could throw away two of $3,6,9$ or throw away one of $1,4,7$ and one of $2,5,8$. 
