a 5 - digit number divisible by 3 is to be formed using numerical 0 , 1 , 2 , 3 , 4 and 5 without repetition . the total number e of ways this can be done is :  Choose the most appropriate option.
A) 122
B) 210
C) 216
D) 217
E) 225

Rationale: 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 these 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 , desired # would be total combinations 5 ! , minus combinations with 0 as the first digit ( combination of 4 ) 4 ! - - > 5 ! - 4 ! = 4 ! ( 5 - 1 ) = 4 ! * 4 = 96 120 + 96 = 216 = e answer : c .
Answer: C