A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is

Answer

216

Explanation

We know that a number is divisible by 3 if the sum of its digits is divisible by 3.
Since the sum of six digits 0,1,2,3,4,5 is 15 which is divisible by 5, we can omit digit either 0 or 3.
So, we have following two cases.
Case I : Digits are 1,2,3,4,5
Here first, second, third, fourth and fifth places can be filled in 5, 4,3,2 and 1 ways respectively.
So, number of numbers in this case are `5xx4xx3xx2xx1=120`
Case II: Digits are 0,1,2,4,5

We have four options to fill first place (1,2,3,4).
For second place again, we have four options as now digit 0 can be considered.
For third, fourth, fifth places we have 3,2 and 1 ways respectively.
So, number of numbers in this case are `4xx4xx3xx2xx1=96`
From cases I and II, total number of numbers=120+96=216.