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 only substracted the numbers with 0 at the peginning however: the divisibilty rule for three says that for a number to be a multiple of 3 the sum of its digits needs to be a multiple of three. So really it does not matter what order you put the numbers in but the sum of the numbers themselves. There are in total 6 combinations of numbers you can pick. which are
$1+2+3+4+5 = 15$
$0+2+3+4+5 = 14$
$0+1+3+4+5 = 13$
$0+1+2+4+5 = 12$
$0+1+2+3+5 = 11$
$0+1+2+3+4 = 10$
Therefore only the combinations that use 1,2,3,4,5 or 0,1,2,3,4 are divisible by 3. There are 5! of the first one = 120. However we cant pick combinations which start with zero so there are 5!-4!of the second one= 96. Add them to get 216.
