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

Options

A. `|ul(10)`
B. `|ul(9)`
C. `9|ul(9)`
D. `10|ul(10)`

Answer

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

### Step 1: Understanding the Digits
We have the digits from 0 to 9, which gives us a total of 10 different digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

### Step 2: Choosing the First Digit
For a 9-digit number, the first digit cannot be 0 (as it would then be an 8-digit number). Therefore, we can choose the first digit from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. This gives us 9 options for the first digit.

### Step 3: Choosing the Remaining Digits
Once we have chosen the first digit, we have 9 digits left (including 0) to choose from for the remaining 8 positions. Since all digits must be different, we will choose the remaining digits as follows:

- For the second digit, we have 9 choices (all digits except the one chosen as the first digit).
- For the third digit, we have 8 choices (all digits except the first two chosen).
- For the fourth digit, we have 7 choices.
- For the fifth digit, we have 6 choices.
- For the sixth digit, we have 5 choices.
- For the seventh digit, we have 4 choices.
- For the eighth digit, we have 3 choices.
- For the ninth digit, we have 2 choices.

### Step 4: Calculating the Total Combinations
The total number of different 9-digit numbers can be calculated by multiplying the number of choices for each digit:

\[ \text{Total Numbers} = 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \]

### Step 5: Simplifying the Expression
This can be simplified as:

\[ \text{Total Numbers} = 9 \times 9! \]

Where \(9!\) (9 factorial) represents the product of all integers from 1 to 9.

### Conclusion
Thus, the total number of 9-digit numbers with all different digits is \(9 \times 9!\).