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

The correct option is B. $18\%$

Given digits are $0,1,2,3,4$ and $5$

For a number to be divisible by $6$ it should be divisible by $2$ as well as $3$.

For divisibility by $2$, last digit should be either $2$ or $4$.

For divisibility by $3$, two cases are suitable:

Case 1:

Discarding '$0$'

Total Cases: $2\times 4!=48$

Case 2:

Discarding '$3$'

Total cases: $4!+2\times 3\times 3!$

Total favourabel cases: $3\times 4!+2\times 3\times 3!=18\times 3!=108$

Sample Space: $5\times 5!$

$P\left(E\right)=\frac{108}{(5\times 5!)}\times 100=\frac{108}{(5\times 120)}\times 100=18\%$