<s>[INST] Want to verify my approach 
If repetition of digit is not allowed, then how many five digit numbers which are divisible by $3$ can be formed
using the digits?

My approach is that i made 3 sets of digits according to the remainder they give on dividing by $3$ -
$ A:(0,3,6,9)$ , $B:(1,4,7)$ , $C:(2,5,8)$, Now the following cases are possible:

*

*Any $1$ digit from $A$ $+$ Any $2$ digit from $B$ $+$ Any $2$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 2} \cdot {3 \choose 2}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 1} \cdot {3 \choose 2} \cdot {3 \choose 2}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$

*Any $2$ digit from $A$ $+$ Any $3$ digit from $B$ $+$ Any $0$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1\cdot {3 \choose 1} \cdot {3 \choose 3} \cdot {3 \choose 0}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 2} \cdot {3 \choose 3}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$

*Any $2$ digit from $A$ $+$ Any $0$ digit from $B$ $+$ Any $3$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 1}\cdot {3 \choose 0} \cdot {3 \choose 3}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 2} \cdot {3 \choose 3}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$

*Any $3$ digit from $A$ $+$ Any $1$ digit from $B$ $+$ Any $1$ digit from $C$ , Now $0$ should not come at Ten thousandths place So handling it in two subcases , when $0$ is chosen from $A$ $\big(1 \cdot {3 \choose 2}\cdot {3 \choose 1} \cdot {3 \choose 1}\cdot4\cdot4\cdot3\cdot2\cdot1 \big) + $ when $0$ not chosen $\big({3 \choose 3} \cdot {3 \choose 1}\cdot {3 \choose 1}\cdot5\cdot4\cdot3\cdot2\cdot1 \big)$.


Total $= 9072$

pls verify my approach and answer, other methods will also be appreciated ,Thanks.
 [/INST] The count can be simplified using probability  as under:
Without zeroes, there will be $\binom9 5 = 126$ digit strings, and the same number with zeroes, at$\,\binom9 4 = 126$
$\frac 1 3$ of each, i.e. $42$ will be divisible by $3$, but since zeroes can only be allowed at $4$ of the $5$ digits, the multiplication factor while combining them won't be $2$ but $1.8$
Thus answer $= 42\times1.8\times5! = 9072\,$, which confirms your answer
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