<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] 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.
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