Q:

Number of such five digit numbers formed of $0,1,2,3,4,5$ which are divisible by $6$ without repetition

I came across the question in the picture on the net.  I couldn't figure
out the solution given.  But what I got is total number of such $5$
digit numbers using the given digits will be $5.5!$. Half of them will
be divisible by $2$. And $1/3$ of them will be divisible by $3$. (Not sure though) What's next?  I couldn't decipher what's written after that...
Please try to use the permutations and combinations approach if possible:

A $5$-digit number is formed by using the digits $0$, $1$, $2$, $3$,
  $4$ & $5$ without repetition.  The probability that the number is
  divisible by $6$ is:
(A) $8\%$
(B) $17\%$
(C) $18\%$
(D) $36\%$
Ans: (C)
Hint: Number should be divisible by $2$ and $3$.
$n(S) = 5 \cdot 5!$; $n(A)$: reject '$0$' $= 2 \cdot 4!$
reject '$3$' $= 4!  + 2 \cdot 3 \cdot 3!$
Total $n(A) = 3 \cdot 4!  + 6 \cdot 3!  = 18 \cdot 3!$
$\therefore p = \frac{18 \cdot 3!}{5 \cdot 5!}  = 18\%$

A:

I. A number $n$ is divisible by $6$ if it is divisible by both $2$ and $3$.
II.  A number $n$ is divisible by $2$ if the rightmost digit is even. 
III.  A number $n$ is divisible by $3$ if the sum of its digits is divisible by $3$. 
We have access to six digits, $\{0,1,2,3,4,5\}$, which sum to $15$. We need a $5$-digit number.  This means we must remove one digit such that the resulting digit-sum is divisible by $3$.
We can do this by either eliminating the $0$ (which results in the digit-sum $15$) or eliminating the $3$ (which results in the digit-sum $12$).  This will allow us to create numbers that are divisible by $3$ no matter how we order the digits.  
Your picture/solution is counting up the possible even numbers for both cases after the divisibility-by-$3$ rule has been satisfied by rejecting a particular digit from the set. 

A:

We try to count how many five-digit numbers formed with digits $0,1,2,3,4,5$ are divisible by six. 
To begin, we note three important facts:

To be a five-digit number, the first digit must not be a zero (otherwise it would technically be a four-digit number.   $01234$ is a four digit number)
To be divisible by six, the number must be divisible by two, so the final digit must be either $0,2,$ or $4$
To be divisible by six, the number must be divisible by three, so the sum of the digits must add up to a multiple of three.   The only way for this to occur with the desired digits and no repetition of digits is if exactly one of $0$ or $3$ is used.   (the proof of which follows from modular arithmetic)

So, putting this information together, we start counting. 
Case 1: $0$ is not used

Pick the final digit: It can either be a two or a four.   - $2$ choices
Pick how the rest of the digits are arranged.   - $4!$ choices

Case 2a: $3$ is not used and $0$ is at the end

Pick how the rest of the digits are arranged.   - $4!$ choices

Case 2b: $3$ is not used and $0$ is not at the end

Pick the final digit: It can be either a two or a four.   -$2$ choices
Pick the location of the zero.   It cannot be at the start or the end.   $3$ choices
Pick how the rest of the digits are arranged.   $3!$ choices

There are then a total of $2\cdot 4!+4!+2\cdot 3\cdot 3!  = 108$ possible five-digit numbers satisfying the conditions. 
The total number of five-digit numbers that can be created with those available digits will be $5\cdot 5!=600$.  (Arrange all six digits in a row, the final digit is unused.   Remove the "bad" arrangements where zero is the leading digit for $6!-5!  = 5\cdot 5!$)
The probability that a five-digit number formed from those digits is divisible by six is then $\frac{108}{600}=0.18$

