How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?

Options

A. 24
B. 120
C. 720
D. None of these

Explanation

To solve the problem of how many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, and 9 that are divisible by 10 with no repeated digits, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Requirement**:
- We need to form a 6-digit number.
- The number must be divisible by 10, which means the last digit must be 0.

2. **Fixing the Last Digit**:
- Since the number must end with 0 (to be divisible by 10), we fix the last digit as 0.
- Our number now looks like this: _ _ _ _ _ 0.

3. **Choosing the First Digit**:
- The first digit cannot be 0 (as it is a 6-digit number), so we can choose from the remaining digits: 1, 3, 5, 7, and 9.
- This gives us 5 options for the first digit.

4. **Filling in the Remaining Digits**:
- After choosing the first digit, we will have 4 digits left to choose from (since one digit has already been used).
- We need to fill in the 4 remaining positions (the second, third, fourth, and fifth digits).
- The number of ways to fill these 4 positions is calculated as follows:
- For the second digit, we have 4 options (the remaining digits).
- For the third digit, we have 3 options (after choosing the second digit).
- For the fourth digit, we have 2 options (after choosing the third digit).
- For the fifth digit, we have 1 option (after choosing the fourth digit).

5. **Calculating the Total Combinations**:
- The total number of combinations can be calculated by multiplying the number of choices at each step:
- Total combinations = (Choices for first digit) × (Choices for second digit) × (Choices for third digit) × (Choices for fourth digit) × (Choices for fifth digit)
- Total combinations = 5 × 4 × 3 × 2 × 1 = 120.

### Final Answer:
Thus, the total number of 6-digit numbers that can be formed under the given conditions is **120**.

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