permutation of 5 digit numbers divisible by 3 "The total number of possible combination of 5 digits numbers formed from the digits(0,1,2,3,4,5,6,7,8,9) which are divisible by 3?"
This was the question given to me by my mathematics teacher during out permutation and combination lessons;I was able to solve this with ease but later I was thought of modifing the problem a little bit to
"The total number of possible combination of 5 digits numbers formed from the digits(0,1,2,3,4,5,6,7,8,9) which are divisible by 3 without repetation of any digits?  eg 12345 not 33120"
I asked my teacher about the solution of the problem because i was unable to solve it with certain accuracy, he was unable to give me a satisfactory answer.
Can anyone help me in solving this problem
thank you.

Here is a correct solution along the lines of Manthanein's answer:
There are $9\cdot8\cdot7\cdot 6  \cdot5=15\,120$ strings of length $5$ not containing a repeat or  zero. Adding $1$ mod $9$ to each digit in such a string  changes its sum by $2$ mod $3$. From this we can conclude  that exactly one third of these strings have a sum which is divisible by $3$. It follows that there are 5040 valid strings of this kind.
Similarly, there are $9\cdot8\cdot7\cdot 6=3024$ strings of length $4$ not containing a repeat or zero, and exactly one third of these strings have a sum which is divisible by $3$. Given such a string we can insert a $0$ at four different places in order to obtain a valid string of length $5$. It follows that there are ${3024\cdot 4\over3}=4032$ valid strings of this kind.
In all there are $5040+4032=9072$ valid strings of length $5$.
