How many 3 digit numbers can be formed by using the digits 1 to 9 if no digit is repeated

Explanation

To solve the problem of how many three-digit numbers can be formed using the digits 1 to 9 without repeating any digit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the available digits**: The digits available for forming the three-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. This gives us a total of 9 digits.

2. **Choose the hundreds place**: For the hundreds place (the first digit), we can choose any of the 9 digits. So, there are 9 options for the hundreds place.

3. **Choose the tens place**: After selecting a digit for the hundreds place, we cannot use that digit again. Therefore, for the tens place (the second digit), we have 8 remaining options (since one digit has already been used).

4. **Choose the units place**: Similarly, after selecting digits for both the hundreds and tens places, we cannot use those two digits again. Thus, for the units place (the third digit), we have 7 remaining options.

5. **Calculate the total combinations**: The total number of different three-digit numbers can be calculated by multiplying the number of options for each place:
\[ \text{Total combinations} = \text{(options for hundreds place)} \times \text{(options for tens place)} \times \text{(options for units place)} \]
\[ \text{Total combinations} = 9 \times 8 \times 7 \]

6. **Perform the multiplication**:
- First, calculate \(9 \times 8 = 72\).
- Then, calculate \(72 \times 7 = 504\).

Thus, the total number of three-digit numbers that can be formed using the digits 1 to 9 without repeating any digit is **504**.