Answer :
Final answer:
Using properties of numbers divisible by 3 and considering the restriction on leading digit, the total number of ways a five-digit number divisible by 3 can be formed from the digits 0, 1, 2, 3, 4, and 5 without repetition is 600. The correct answer is option c.
Explanation:
To find the total number of ways a five-digit number can be formed from the digits 0, 1, 2, 3, 4, and 5 without repetition and divisible by 3, we first need to consider the properties of numbers divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. We have six digits, which sum up to 0 + 1 + 2 + 3 + 4 + 5 = 15, a number divisible by 3, meaning any combination of these digits without repetition will also result in a number divisible by 3.
However, we need to be careful with the digit '0' as it cannot be the leading digit. There are 5 options for the first digit (1, 2, 3, 4, or 5), followed by any one of the remaining 5 digits for the second position, 4 for the third, 3 for the fourth, and 2 for the last which includes '0' since it can now be used. So the formula for the permutation without repeats would yield the total ways as 5 * 5 * 4 * 3 * 2 = 600 ways.
