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: 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
A: Your second method does not take in to account that the numbers must be divisible by 3. If you had consecutive numbers you could divide by 3, which also seems to work here, but I suspect the best approach is to determine how many triples of {1,2,3,4,5} sum to a multiple of 3, and use that as your set instead of using the entire set, without 2 or 4.
A: Hint:-Form even numbers such that their sum is divisible by 3.
Divisibility rule by 6 says that the number must be divisible by both 2 and 3.Your second try does not take into account that the number must be divisible by 3.The possible ways are ,as you quoted-(1,3,2)
(3,1,2)
(2,3,4)
(3,2,4)
(4,3,2)
(3,4,2)
(3,5,4)
(5,3,4).There is possibly no better way to do it without using divisibility.
