Permutation and Combination with divisibility? How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?
my explanation:
total number of permutations with 0, 1, 2, 3, 4 and 5 
to have 5 digits = 6.5.4.3.2=720 (includes numbers with 0 at the beginning)
now let us find the numbers with zero at the beginning = 5.4.3.2=120
so total no of digits = 720 - 120 = 600
but this is not the answer? the correct answer is 216.
can somebody correct me whats wrong with my approach and suggest a better solution.

You haven’t taken into account the requirement that the numbers be divisible by $3$.
Since you’re allowed to use only the six digits $0,1,2,3,4$, and $5$ and are not allowed to repeat any digit, there are just $6$ possible sets of digits that you can use, $\{0,1,2,3,4\}$, $\{0,1,2,3,5\}$, $\{0,1,2,4,5\}$, $\{0,1,3,4,5\}$, $\{0,2,3,4,5\}$, and $\{1,2,3,4,5\}$. An integer is divisible by $3$ if and only if the sum of its digits is divisible by $3$, so only the sets $\{0,1,2,4,5\}$ and $\{1,2,3,4,5\}$ will actually give you multiples of $3$. Every permutation of each of these sets gives you a multiple of $3$.


*

*How many permutations of the digits $1,2,3,4$, and $5$ are there?  

*How many permutations of the digits $0,1,2,4$, and $5$ are there that do not start with $0$?


Thus sum of those two numbers is what you’re looking for.
