Statement-1: A 5-digit number divisible by 3 is to be formed using the digits 0,1,2,3,4,5 without repetition, then the total number of ways this can be done is 216. Statement-2: A number is divisible by 3, if sum of its digits is divisible by 3.

Options

A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1
C. Statement-1 is true, statement-2 is false
D. Statement-1 is false, statement-2 is true

Explanation

To solve the problem, we need to determine the total number of 5-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition, such that the number is divisible by 3.

### Step-by-Step Solution:

1. **Understanding the Digits**: The available digits are 0, 1, 2, 3, 4, and 5. We need to form a 5-digit number using these digits without repetition.

2. **Condition for Divisibility by 3**: A number is divisible by 3 if the sum of its digits is divisible by 3. We will first calculate the sum of all available digits:
\[ 0 + 1 + 2 + 3 + 4 + 5 = 15 \]
Since 15 is divisible by 3, any combination of 5 digits from this set will also be divisible by 3 if the excluded digit is also divisible by 3.

3. **Identifying Excluded Digits**: The digits that are divisible by 3 from our set are 0 and 3. Thus, we can exclude either 0 or 3 to ensure the sum of the remaining digits is divisible by 3.

4. **Case 1: Excluding 0**:
- The remaining digits are 1, 2, 3, 4, and 5.
- We can form a 5-digit number using all these digits. The total number of arrangements is:
\[ 5! = 120 \]

5. **Case 2: Excluding 3**:
- The remaining digits are 0, 1, 2, 4, and 5.
- Since we are forming a 5-digit number, 0 cannot be the leading digit. Therefore, we have to consider the arrangements where 0 is not in the first position.
- We can choose any of the digits 1, 2, 4, or 5 as the first digit (4 choices), and then arrange the remaining 4 digits (including 0) in the remaining 4 positions:
\[ 4 \times 4! = 4 \times 24 = 96 \]

6. **Total Combinations**: Now, we add the two cases together:
\[ 120 \text{ (from Case 1)} + 96 \text{ (from Case 2)} = 216 \]

7. **Conclusion**: The total number of 5-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition, such that the number is divisible by 3, is indeed 216.

### Final Statements:
- Statement 1 is TRUE.
- Statement 2 is TRUE, as it correctly explains the condition for divisibility by 3.