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?
 A: 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.
A: Case $1$:
$1,2,3,4,5$ are chosen.
For the number to be divisible by $2$, last digit is $2$ or $4$, i.e. last digit can be selected in 2 ways and the other digits can arrange in $4!$. Thus, the number of ways =$4! \cdot 2=48$.
Case $2$:
$0,1,2,4,5$ are chosen.
a) if last digit is $0$: number of ways =$4!=24$
b) if last digit is $2$: the other digits may be filled in (from left to right): $3×3×2×1=18$ ways.
c) if last digit is $4$: the other digits may be filled in (from left to right): $3×3×2×1=18$ ways.
So, the answer $=48+24+18+18=108$ ways.
So answer is $108$.
A: The sum of the digits must be divisible by $3$, so either $0$ or $3$ must be absent (since $0+1+2+3+4+5=15$ is divisible by $3$).
If $0$ is absent, then we have a permutation of $12345$ where the last digit is $2$ or $4$, giving $(2)(4!)=(2)(24)=48$ possibilities.
If $3$ is absent, then we have a permutation of $01245$ where the first digit is not $0$ and the last digit is $0, 2,$ or $4$. Such permutations could be counted as follows:
First, consider the case where the last digit is $0$. In this case, the remaining digits must form a permutation of $1245$, giving $4!=24$ possibilities.
Second, consider the case where the last digit is $2$ or $4$. In this case, the remaining digits must form a permutation of $0145$ (if the last digit is $2$) or $0125$ (if the last digit is $4$) where the first digit is not $0$. This gives $(2)(4!-3!)=(2)(24-6)=(2)(18)=36$ possibilities.
Thus, there are $48+24+36=108$ possible numbers.
