A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done is A. 216 B. 600 C. 240 D. 3125 [Hint:5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.]

To solve this problem, we need to consider two cases:

1. The five-digit number is formed using the digits 0, 1, 2, 4, 5. Since 0 cannot be the first digit, we have 4 options for the first digit, 4 options for the second digit (including 0), 3 options for the third digit, 2 options for the fourth digit, and 1 option for the fifth digit. So, the total number of ways in this case is 44321 = 96.

2. The five-digit number is formed using the digits 1, 2, 3, 4, 5. Since all these digits can be the first digit, we have 5 options for the first digit, 4 options for the second digit, 3 options for the third digit, 2 options for the fourth digit, and 1 option for the fifth digit. So, the total number of ways in this case is 54321 = 120.

Adding the number of ways in both cases, we get 96 + 120 = 216.

So, the total number of ways to form a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, 5 without repetitions is 216. Therefore, the answer is A. 216.