<s>[INST] 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?
 [/INST] A refinement of your first approach:

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


*

*$132$ the sum of the other two digits is 4

*$312$ the sum of the other two digits is 4

*$342$ the sum of the other two digits is 7

*$432$ the sum of the other two digits is 7

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


*

*$234$ the sum of the other two digits is 5

*$324$ the sum of the other two digits is 5

*$354$ the sum of the other two digits is 8

*$534$ the sum of the other two digits is 8
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