How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once?

The total number of $5$ digit numbers using the digits $0,1,2,3,4,7$ and $8$ is $6\times6\times5\times4\times3=2160.$  
Now I found the numbers not divisible by $3$, i.e. even numbers ending in $2,4,8.$  
Even numbers from the digits $0,1,2,3,4,7$ and $8$ are $5\times5\times4\times3\times3=930.$  
So the numbers divisible by $3$ are $2160-930=1230$ but the answer is $744.$ Where I am wrong?

Solution
We've to make $5$ digit numbers using given $7$ digits excluding $2$ digits every time.
Now,$0+1+2+3+4+7+8=25$.
As $25 \pmod 3 \equiv 1$, duplets to be excluded should have their sum $\mod 3=1$.
It's not much difficult to find such duplets. Without much trouble,
we get :$$(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7).$$
Thus, there are such $7$ duplets & hence $7$ such groups of $5$ digits.
Now, we find that out of $7$ groups, $4$ groups have $0$ in them which isn't allowed to take 5th place ( otherwise a $4$-digit number will be formed.).
Total numbers formed by these $4$ groups $=4(5!−4!)=384$.
Other $3$ groups have no zero so total numbers formed by them$=3(5!)=360$
Total possibel numbers: $384+360=744$
