A five digit number divisible by 3 is to be formed using the 

numericals 0, 1, 2, 3, 4 and 5, without repetition. The total

 number of ways this can be done, is 

Solution:

If a number is divisible by 3, the sum of the digits

 in it must be a multiple of 3. The sum of the given 

six numerals is $0 + 1 + 2 + 3 + 4 + 5 = 15$. So, to make 

a five digit number divisible by 3 we can either exclude

 0 or 3. If 0 is left out, then 5! = 120 number of ways are

 possible. If 3 is left out, then the number of ways of making

 a five digit number is $4 x 4 ! = 96$ , because 0 cannot be placed 

in the first place from left, as it will give a number of four digits. 

 Thus, the required number of ways $= 120 + 96 = 216.$