The total number of 9-digit number which have all different digits is

Options

A. `10"!"`
B. `9"!"`
C. `9 xx 9"!"`
D. `10 xx 10"!"`

Explanation

To find the total number of 9-digit numbers that have all different digits, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Digits**:
- A 9-digit number can only use the digits 1 to 9 and can include the digit 0. However, since it is a 9-digit number, the first digit cannot be 0. Therefore, we have 9 possible digits (1 to 9) for the first position.

2. **Choosing the First Digit**:
- For the first digit, we have 9 options (1 to 9). We cannot use 0 here because it would make it an 8-digit number.

3. **Choosing the Remaining Digits**:
- After choosing the first digit, we have 9 digits left (including 0) to fill the remaining 8 positions.
- For the second digit, we can choose from the remaining 9 digits (including 0).
- For the third digit, we can choose from the remaining 8 digits.
- This pattern continues until we fill all 9 positions.

4. **Calculating the Total Combinations**:
- The total number of ways to arrange these digits can be calculated as follows:
- First digit: 9 choices (1-9)
- Second digit: 9 choices (remaining digits including 0)
- Third digit: 8 choices (remaining digits)
- Fourth digit: 7 choices
- Fifth digit: 6 choices
- Sixth digit: 5 choices
- Seventh digit: 4 choices
- Eighth digit: 3 choices
- Ninth digit: 2 choices
- Therefore, the total number of 9-digit numbers with all different digits is:
\[ 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \]

5. **Simplifying the Calculation**:
- This can also be expressed as:
\[ 9 \times 9! \quad \text{(where } 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\text{)} \]
- Thus, the total number of 9-digit numbers with all different digits is:
\[ 9 \times 9! \]

### Final Answer:
The total number of 9-digit numbers that have all different digits is \( 9 \times 9! \).