Five digit number divisible by $$3$$ is formed using $$0, 1, 2, 3, 4$$   and $$5$$ without repetition. Total number of such numbers are

Solution :
We know that a number is divisible by 3 only when the sum of the digits is divisible by 3. The given digits are $$0, 1, 2, 3, 4, 5 .$$
Here the possible number of combinations of $$5$$ digits out of $$6$$ are $$^5{C_4} = 5,$$   which are as follows -
$$1 + 2 + 3 + 4 + 5 = 15 = 3 \times 5$$
$$0 + 2 + 3 + 4 + 5 = 14$$     (not divisible by $$3$$ )
$$0 + 1 + 3 + 4 + 5 = 13$$     (not divisible by $$3$$ )
$$0 + 1 + 2 + 4 + 5 = 12 = 3 \times 4$$
$$0 + 1 + 2 + 3 + 5 = 11$$     (not divisible by $$3$$ )
$$0 + 1 + 2 + 3 + 4 = 10$$     (not divisible by $$3$$ )
Thus the number should contain the digits $$1, 2, 3, 4, 5$$   or the digits $$0, 1,2, 4, 5.$$
Taking $$1, 2, 3, 4, 5,$$   the $$5$$ digit numbers are $$= 5! = 120$$
Taking $$0, 1, 2, 4, 5,$$   the $$5$$ digit numbers are $$= 5! - 4! = 96$$
∴ Total number of numbers $$= 120 + 96 = 216$$