The correct option is
C 18%According to the question,
Total 5 digit numbers formed by the given digits=5∗5!
For divisibility by 6, the numbers must be divisible by both 2 and 3.
Only numbers formed by digits
(i) Case 1: (0,1,2,4,5) For divisibility by 2,......
If 0 is placed at the end, the numbers of ways=4!=24
If 0 is not place din the end, then their are two ways to select the last digit$(2,4).
Number of ways=2C1∗3C1∗3! =36
For this case total number of ways=24+36=60
(ii) Case 2: (1,2,3,4,5) Again, the last number can be chosen in 2C1ways and the remaining 4 numbers can permute in 4! ways.
The required number of ways for this case=2∗4!=48
Adding the number of ways in the two cases, 5 digit numbers that are divisible by 6=108$
Required Probability108600=0.18
=18%