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.