To solve the problem of finding the total number of 10-digit numbers that can be formed using the digits 1 to 9, where one digit is repeated, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Problem**:
We need to create a 10-digit number using the digits 1 to 9, with one of these digits appearing twice.
2. **Choosing the Repeated Digit**:
We can choose any one of the 9 digits (1 to 9) to be the repeated digit. The number of ways to choose one digit from these 9 is given by \( \binom{9}{1} \), which is 9.
\[
\text{Ways to choose the repeated digit} = 9
\]
3. **Arranging the Digits**:
After choosing the repeated digit, we will have 10 digits in total: 9 different digits (1 to 9) and one of them repeated.
The arrangement of these 10 digits can be calculated using the formula for permutations of multiset:
\[
\text{Total arrangements} = \frac{10!}{2!}
\]
Here, \( 10! \) is the factorial of the total number of digits, and \( 2! \) accounts for the repetition of one digit.
4. **Calculating the Total Number of Arrangements**:
Now, we can combine the choices and arrangements:
\[
\text{Total numbers} = \text{Ways to choose the repeated digit} \times \text{Total arrangements}
\]
Substituting the values we calculated:
\[
\text{Total numbers} = 9 \times \frac{10!}{2!}
\]
5. **Final Calculation**:
We know that \( 2! = 2 \), so we can simplify:
\[
\text{Total numbers} = 9 \times \frac{10!}{2} = \frac{9 \times 10!}{2}
\]
### Conclusion:
Thus, the total number of 10-digit numbers in which only and all the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 appear, with one digit repeated, is:
\[
\frac{9 \times 10!}{2}
\]
