a 5 digit number divisible by 3 is to be formed using the digits 0 , 1 , 2 , 3 , 4 and 5 without repetitions . that total no of ways it can be done is ?  Options.
A) 122
B) 210
C) 216
D) 217
E) 220

Rationale: first step : we should determine which 5 digits from given 6 , would form the 5 digit number divisible by 3 . we have six digits : 0,1 , 2,3 , 4,5 . their sum = 15 . for a number to be divisible by 3 the sum of the digits must be divisible by 3 . as the sum of the six given numbers is 15 ( divisible by 3 ) only 5 digits good to form our 5 digit number would be 15 - 0 = ( 1 , 2,3 , 4,5 ) and 15 - 3 = ( 0 , 1,2 , 4,5 ) . meaning that no other 5 from given six will total the number divisible by 3 . second step : we have two set of numbers : 1 , 2,3 , 4,5 and 0 , 1,2 , 4,5 . how many 5 digit numbers can be formed using this two sets : 1 , 2,3 , 4,5 - - > 5 ! as any combination of these digits would give us 5 digit number divisible by 3 . 5 ! = 120 . 0 , 1,2 , 4,5 - - > here we can not use 0 as the first digit , otherwise number wo n ' t be any more 5 digit and become 4 digit . so , total combinations 5 ! , minus combinations with 0 as the first digit ( combination of 4 ) 4 ! - - > 5 ! - 4 ! = 96 120 + 96 = 216 answer : c .
Answer: C