How many 5-digit numbers can be formed from the integers 1,2, …, 9, if no digit can appear more than twice?

We are tasked with finding how many 5-digit numbers can be formed from the integers 1, 2, …, 9, under the condition that no digit can appear more than twice.

### Step 1: Understanding the Restrictions
- We can only use the digits \( 1, 2, \dots, 9 \), so there are 9 possible digits to choose from.
- No digit can appear more than twice.

### Step 2: Total Number of Positions
A 5-digit number has 5 positions to fill. Let's call these positions \( d_1, d_2, d_3, d_4, d_5 \).

### Step 3: Choose Digits and Their Frequency
The digits in the number must meet the condition that no digit appears more than twice. This means that, for any 5-digit number:
- Some digits can appear twice.
- The rest of the digits can appear only once.

We can split the number of occurrences of the digits into two cases:
- **Case 1**: Two digits appear twice, and one digit appears once.
- **Case 2**: One digit appears twice, and three digits appear once.

#### Case 1: Two Digits Appearing Twice, One Digit Appearing Once
1. **Choose two digits to appear twice**: We can choose 2 digits from the 9 available digits. This can be done in \( \binom{9}{2} = \frac{9 \times 8}{2} = 36 \) ways.
2. **Choose one digit to appear once**: From the remaining 7 digits, choose 1 to appear once. This can be done in \( \binom{7}{1} = 7 \) ways.
3. **Arrange the digits**: We need to arrange the chosen digits (where two appear twice and one appears once) into the 5 positions. This can be done as follows:
   - First, choose 2 positions out of 5 for the first digit that appears twice, which can be done in \( \binom{5}{2} = 10 \) ways.
   - Next, choose 2 positions out of the remaining 3 for the second digit that appears twice, which can be done in \( \binom{3}{2} = 3 \) ways.
   - Finally, the remaining position must be filled by the last digit, so there is only 1 way to arrange this.

Thus, the total number of ways for **Case 1** is:
\[
36 \times 7 \times 10 \times 3 \times 1 = 7560.
\]

#### Case 2: One Digit Appearing Twice, Three Digits Appearing Once
1. **Choose one digit to appear twice**: We can choose 1 digit from the 9 available digits to appear twice. This can be done in \( \binom{9}{1} = 9 \) ways.
2. **Choose three digits to appear once**: From the remaining 8 digits, choose 3 to appear once. This can be done in \( \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \) ways.
3. **Arrange the digits**: We need to arrange the chosen digits (where one appears twice and three appear once) into the 5 positions. This can be done as follows:
   - First, choose 2 positions out of 5 for the digit that appears twice, which can be done in \( \binom{5}{2} = 10 \) ways.
   - The remaining 3 positions must be filled by the digits that each appear once, which can be done in \( 3! = 6 \) ways.

Thus, the total number of ways for **Case 2** is:
\[
9 \times 56 \times 10 \times 6 = 30240.
\]

### Step 4: Total Number of 5-Digit Numbers
The total number of 5-digit numbers is the sum of the results from both cases:
\[
7560 + 30240 = 37800.
\]

Thus, the total number of 5-digit numbers that can be formed from the digits 1, 2, ..., 9, such that no digit appears more than twice, is **37,800**.

Would you like further details or have any questions?

Here are 5 related questions to explore:
1. How many 6-digit numbers can be formed if no digit can appear more than twice?
2. What if the digits can appear up to three times in a 5-digit number?
3. How many 5-digit numbers can be formed if repetition of digits is not allowed?
4. What is the probability of randomly selecting a number where a specific digit appears exactly twice?
5. How does the answer change if the digits include 0?

**Tip:** When dealing with restrictions on digit appearances, it's helpful to break down the cases based on how many times digits can appear in each position.

How many 5-digit numbers can be formed from the integers 1, 2, …, 9, if no digit can appear more than twice?

Find out how many 5-digit numbers can be created using the digits 1-9 with no digit appearing more than twice. This problem uses combinatorics, including combination and permutation formulas, and is suitable for high school students in Grades 10-12.

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