Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed. 
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.

I started by considering the following cases:

*

*Unit digit = $0$
I can fill four places with digits ${1,2,4,5}$ , so number of 5-digit numbers $= P(4,4) = 24$

*Unit digit = $2$
I can fill four places with digits ${0,1,3,4,5}$ such that (a) $0$ does not come in the first place (b) either of $1,4$  is used (c) $0,3,5$ are always used. I have to fill 4 places with $0,(1,4),3,5$. So, total number of arrangements $=P(4,4)$. Number of arrangements when $0$ comes as first digit $=P(3,3)$. Number of arrangements of $(1,4) = 2$. Therefore , number of 5-digit numbers $=2(P(4,4)-P(3,3)) = 36$

*Unit digit = $4$
Similarly, number of 5-digit numbers $= 36$
Therefore, the total number of 5-digit numbers $= 24+36+36 = 96$.
But the correct answer is $108$. Where did I make a mistake?

The number has to be divisible by $3$, so the sum of the digits must also have that property.  Since the sum of the six digits is $15$, the missing digit has to be either $0$ or $3$.  (This was your mistake -- you thought that either 1 or 4 needed to be unused).

*

*If the missing digit is $0$, then the units digit can be chosen in two ways and the remaining four digits can be arranged in $4!=24$ ways.  This leads to $2\cdot24=48$ cases.

*If the missing digit is $3$ and $0$ is the units digit, then the remaining four digits can be arranged in $4!=24$ ways,

*If the missing digit is $3$ and $0$ is not the units digit, then the units digit can be chosen in two ways.  From the remaking four digits, the ten-thousands digit can be chosen in three ways (to avoid a leading zero), and the remaining three digits can be arranged in $3!=6$ ways.  This leads to $2\cdot3\cdot6=36$ cases.

The total is $48+24+36=108$ possible cases.
