Understanding the Problem Range

The question asks for the number of integers lying between 2999 and 8001. This range includes all integers strictly greater than 2999 and strictly less than 8001. Therefore, the integers we are considering are 3000, 3001, 3002, ..., up to 8000. This range can be represented as [3000, 8000].

Total Integers in the Range

The total number of integers in the range from 3000 to 8000 (inclusive) can be calculated as:

Total number of integers = Ending number - Starting number + 1

Total number of integers = $8000 - 3000 + 1 = 5001$

So, there are 5001 integers between 2999 and 8001.

Strategy: Using the Complement Approach

We are looking for integers that have at least two digits equal. It is often easier to calculate the opposite case: the number of integers where all digits are distinct. Once we have this number, we can subtract it from the total number of integers in the range to find the number of integers with at least two digits equal (i.e., repeated digits).

Numbers with at least two digits equal = Total numbers - Numbers with all distinct digits

Counting Integers with Distinct Digits (3000 to 7999)

The integers in the range [3000, 8000] are mostly 4-digit numbers. The only exception is 8000, which is a 4-digit number but needs special consideration due to its leading digit being 8.

Let's first count the number of 4-digit integers in the range [3000, 7999] that have all distinct digits. A 4-digit number can be represented as $d_1 d_2 d_3 d_4$.

• The first digit ($d_1$) must be between 3 and 7 (inclusive) for the number to be in the range [3000, 7999]. So, $d_1$ can be 3, 4, 5, 6, or 7. There are 5 choices for the first digit.
• The second digit ($d_2$) can be any digit from 0 to 9, but it must be different from $d_1$ (since digits must be distinct). There are 10 total digits (0-9). After choosing $d_1$, there are $10 - 1 = 9$ digits remaining. So, there are 9 choices for the second digit.
• The third digit ($d_3$) can be any digit from 0 to 9, but it must be different from $d_1$ and $d_2$. After choosing $d_1$ and $d_2$, there are $10 - 2 = 8$ digits remaining. So, there are 8 choices for the third digit.
• The fourth digit ($d_4$) can be any digit from 0 to 9, but it must be different from $d_1$, $d_2$, and $d_3$. After choosing $d_1$, $d_2$, and $d_3$, there are $10 - 3 = 7$ digits remaining. So, there are 7 choices for the fourth digit.

The total number of integers in the range [3000, 7999] with all distinct digits is the product of the number of choices for each digit:

Number of distinct digit integers in [3000, 7999] = $5 \times 9 \times 8 \times 7$

Calculation:

$5 \times 9 = 45$

$45 \times 8 = 360$

$360 \times 7 = 2520$

So, there are 2520 integers between 3000 and 7999 that have all distinct digits.

Calculating Integers with Repeated Digits (3000 to 7999)

The total number of integers in the range [3000, 7999] is $7999 - 3000 + 1 = 5000$.

The number of integers in [3000, 7999] with at least two digits equal is:

Number of repeated digit integers in [3000, 7999] = Total numbers in [3000, 7999] - Number of distinct digit integers in [3000, 7999]

Number of repeated digit integers in [3000, 7999] = $5000 - 2520 = 2480$

Analyzing the Number 8000

The range of integers is [3000, 8000]. We have considered the integers from 3000 up to 7999. Now we need to consider the number 8000 itself.

The digits of 8000 are 8, 0, 0, 0. This number has multiple digits equal (the digit 0 appears three times). Therefore, the number 8000 has at least two digits equal.

Combining Results for the Full Range (3000 to 8000)

To find the total number of integers in the range [3000, 8000] with at least two digits equal, we add the count from the range [3000, 7999] and the count for 8000.

• Number of integers in [3000, 7999] with at least two digits equal = 2480
• Does 8000 have at least two digits equal? Yes. So, count = 1.

Total number of integers in [3000, 8000] with at least two digits equal = (Repeated digit integers in [3000, 7999]) + (Check for 8000)

Total = $2480 + 1 = 2481$

Final Answer Summary

The number of integers lying between 2999 and 8001 (inclusive of 3000 and 8000) which have at least two digits equal is 2481.

Revision Table: Key Counting Concepts

Additional Information: Exploring Number Properties

Problems involving counting numbers with specific digit properties are common in combinatorics and number theory. The complement principle is a powerful tool in such cases, especially when dealing with "at least one" conditions.

The concept of permutations is used when the order of digits matters, as it does in forming distinct numbers. If the problem involved sets of digits where order didn't matter, combinations might be used.

The Pigeonhole Principle is also relevant in a broader sense to digit problems. For example, if you have a number with more than 10 digits, at least one digit must be repeated because there are only 10 possible digits (0-9).

Variations of this problem could involve different ranges (e.g., including leading zeros if discussing strings), different number of digits, or different conditions on the digits (e.g., digits must be even, sum of digits is fixed, etc.). Each variation requires careful consideration of the constraints and available counting techniques.