Problem:
A five-digit number divisible by 3 is to be formed using numerals 0, 1, 2, 3, 4, and 5 without any repetition. The total number of ways in which this can be done is _______.

Solution:

To find the total number of ways to form a five-digit number divisible by 3 using the given numerals without repetition, we need to consider a few key points:

Key Points:
1. A number is divisible by 3 if the sum of its digits is divisible by 3.
2. The digit 0 cannot be used as the leading digit in the five-digit number.
3. The digit 0 can be used as one of the non-leading digits.

Approach:
To form a five-digit number divisible by 3:
1. We first select the leading digit from the given numerals (1, 2, 3, 4, 5) except 0.
2. We then select four more digits from the remaining numerals (0, 1, 2, 3, 4, 5) without repetition.
3. We arrange these five digits in different orders to form a five-digit number.
4. We check if the sum of these five digits is divisible by 3.
5. If it is divisible by 3, we count this arrangement as one valid way.

Calculation:
Let's analyze the possible cases:

Case 1: Leading digit: 1
- Remaining digits: 0, 2, 3, 4, 5
- Total ways to select 4 digits without repetition: 5P4 = 5! / (5 - 4)! = 5 * 4 * 3 * 2 = 120
- Total ways to arrange these 5 digits: 5!
- Total valid ways: Number of ways to select 4 digits without repetition * Number of ways to arrange these 5 digits

Case 2: Leading digit: 2
- Remaining digits: 0, 1, 3, 4, 5
- Total ways to select 4 digits without repetition: 5P4 = 5! / (5 - 4)! = 5 * 4 * 3 * 2 = 120
- Total ways to arrange these 5 digits: 5!
- Total valid ways: Number of ways to select 4 digits without repetition * Number of ways to arrange these 5 digits

Case 3: Leading digit: 3
- Remaining digits: 0, 1, 2, 4, 5
- Total ways to select 4 digits without repetition: 5P4 = 5! / (5 - 4)! = 5 * 4 * 3 * 2 = 120
- Total ways to arrange these 5 digits: 5!
- Total valid ways: Number of ways to select 4 digits without repetition * Number of ways to arrange these 5 digits

Case 4: Leading digit: 4
- Remaining digits: 0, 1, 2, 3, 5
- Total ways to select 4 digits without repetition:

The sum of 0, 1, 2, 3, 4 and 5 is 15. For the numbers to be divisible by 3, the sum of the digits should be a multiple of 3. We can have five-digits numbers by removing either 0 or 3 so that the number is divisible by 3. If we remove 0, we can have 5! = 120 five-digit numbers without repetitions. If we remove 3, we can have 4! ×4 = 96 five-digit numbers. So the total numbers possible = 120 + 96 = 216.