To find the total number of 10-digit numbers that can be formed using all the digits from 1 to 9, with one digit repeated, we can follow these steps:
### Step 1: Understand the problem
We need to create a 10-digit number using the digits 1 to 9, where each digit from 1 to 9 appears at least once, and one of these digits must be repeated.
### Step 2: Identify the digits
The digits available are: 1, 2, 3, 4, 5, 6, 7, 8, 9. There are a total of 9 different digits.
### Step 3: Choose the digit to be repeated
Since we need to repeat one of the digits, we can choose any one of the 9 digits to repeat. The number of ways to choose the digit to be repeated is given by:
\[
\text{Number of ways to choose the digit to repeat} = \binom{9}{1} = 9
\]
### Step 4: Arrange the digits
After choosing the digit to be repeated, we will have 10 digits in total: 9 different digits and 1 digit repeated. The arrangement of these 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 the chosen digit.
### Step 5: Calculate the total number of arrangements
Now, we can calculate the total number of 10-digit numbers:
\[
\text{Total numbers} = \text{Number of ways to choose the digit to repeat} \times \text{Total arrangements}
\]
Substituting the values we found:
\[
\text{Total numbers} = 9 \times \frac{10!}{2!}
\]
### Step 6: Final calculation
Now we can compute the final answer:
\[
\text{Total numbers} = 9 \times \frac{10!}{2} = \frac{9 \times 10!}{2}
\]
### Conclusion
Thus, the total number of 10-digit numbers in which all digits from 1 to 9 appear, with one digit repeated, is:
\[
\frac{9 \times 10!}{2}
\]
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