A 5 digit number is formed using the digits 0,1,2,3,4 and 5, probability it is divisible by 6? 
A $5$ digit number is formed by using the digits $0,1,2,3,4$ and $5$ without
  repetition. The probability that the number is divisible by $6$ is?

Answer : 18%
I had doubts regarding this question but while writing my attempt here, I got the solution. So, I am posting it as an answer. 


Total $5$ digit numbers formed by the given digits=$5*5!=600$

For divisibility by 6, the numbers must be divisible by both 2 and 3. 
Only numbers formed by digits (i) (0,1,2,4,5) and (ii)
    (1,2,3,4,5) are divisible by 3.
(i)  CASE 1: ( 0,1,2,4,5)   For divisibility by 2, _ _ _ _ _ 
(a) if 0 is placed at the end, then number of ways=4!=24   (b) if 0 is not placed in the end, then there are two ways to select the last
    digit(2,4). Number of ways=$2C1*3C1*3!=36$
    For this case, total number of ways=24+36=60
(ii) CASE 2: ( 1,2,3,4,5)    Again, the last number can be chosen in $2C1$ ways and the
    remaining 4 numbers can permute in $4!$ ways. The required number of
    ways for this case= $2*4!= 48$
Adding the number of ways in the two cases, 5 digit numbers that are
    divisible by 6= 108
Required Probability = $\frac{108}{600} = $0.18$= $18%


