[Math] How many 3 digit numbers can be formed using digits 1,2,3,4 and 5 such that the number is divisible by 6

How many $3$ digit numbers can be formed using digits $1,2,3,4$ and $5$ without repetition such that the number is divisible by $6$

First Approach:

A number is divisible by $6$ if it is divisible by $2$ and $3$.

Now the possible combinations I found are

$(1,3,2)$

$(3,1,2)$

$(2,3,4)$

$(3,2,4)$

$(4,3,2)$

$(3,4,2)$

$(3,5,4)$

$(5,3,4)$

total $8$ ways.

Second Approach

Case1:
unit digit can be filled in only two ways $(2,4)$ for nos $(3,2,4)$

Tens digit can be filled in $2$ ways

Hundred digit can be filled in $1$ ways

the required number is 2*1*2=4 ways

Case2:
unit digit can be filled in only one ways $(2)$ for nos $(1,2,3)$

Tens digit can be filled in $1$ ways

Hundred digit can be filled in $2$ ways

the required number is 2*1*1=2 ways

Case3:
unit digit can be filled in only one ways $(4)$ for nos $(3,4,5)$

Tens digit can be filled in $1$ ways

Hundred digit can be filled in $2$ ways

the required number is 2*1*1=2 ways

So, Total ways=8

Is there still a better way to solve this problem?

A refinement of your first approach:

In numbers which end with $2$, the sum of the other two digits must be $4$ or $7$:

In numbers which end with $4$, the sum of the other two digits must be $5$ or $8$: