A five-digit numbers 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

Solution :
KEY CONCEPT : We know that a number is divisible by 3 if the sum of its digits is divisibly by 3.
Now out of 0, 1, 2, 3, 4, 5 if we take 1, 2, 3, 4, 5 or 0 , 1, 2, 4, 5 then the 5 digit numbers will be divisible by 3.
Case I : Number of 5 digit numbers formed using the digits 1, 2, 3, 4, 5 = 5! = 120
Case II : Taking 0, 1, 2, 4, 5 if we make 5 digit number then
I place can be filled in = 4 ways (0 can not come at I place)
II place can be filled in = 4 ways
III place can be filled in = 3 ways
IV place can be filled in = 2 ways
V place can be filled in = 1 ways
∴ Total numbers are $$ = 4 \times 4! = 96$$
Thus total numbers divisible by 3 are = 120 + 96 = 216