How many integers between 1 and 10^6 satisfy the condition that the sum of their digits is equal to 9?

We need to find how many integers between \(1\) and \(10^6\) satisfy the condition that the sum of their digits is equal to \(9\).

An integer between \(1\) and \(10^6\) has up to 6 digits. We will treat this problem as finding how many 6-digit numbers (with leading zeros allowed) have their digits summing to 9.

### Approach:
1. **Interpret the problem as a combinatorial problem**: 
   We need to count how many ways we can assign non-negative digits (from 0 to 9) to six positions such that their sum equals 9.

2. **Formulate it as a stars and bars problem**:
   Let the six digits of the number be \(d_1, d_2, \dots, d_6\), where \(d_1 + d_2 + \dots + d_6 = 9\) and \(0 \leq d_i \leq 9\) for each \(i\).
   
   The number of solutions to this equation, where the digits can be any non-negative integer, would be a typical "stars and bars" problem: distributing 9 units (stars) among 6 groups (digits). This is calculated using the binomial coefficient:
   \[
   \binom{9+6-1}{6-1} = \binom{14}{5}.
   \]
   This counts all ways of distributing 9 units among 6 positions, without restriction on the size of each position.

3. **Account for the digit constraint**: 
   The digits \(d_1, d_2, \dots, d_6\) are restricted to being between 0 and 9, so we need to ensure no digit exceeds 9. However, since the total sum is 9, which is small compared to 6 digits where each can take values up to 9, no digit can exceed 9 in this case. Thus, no further adjustment is needed.

### Final Calculation:
We compute the binomial coefficient:
\[
\binom{14}{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002.
\]

Thus, there are **2002 integers** between 1 and \(10^6\) whose digits sum to 9.

Would you like more details or have any further questions?

Here are 5 related questions to explore further:
1. How would the solution change if the sum of the digits were 15 instead of 9?
2. What if we considered only 5-digit numbers instead of 6-digit numbers?
3. How many integers between 1 and \(10^6\) have their digits summing to a value less than 9?
4. What is the probability that a randomly chosen number between 1 and \(10^6\) has its digits summing to 9?
5. How does the stars and bars method work in general for problems involving partitions of numbers?

**Tip:** When solving combinatorial problems, consider if there are restrictions or boundaries that may limit certain solutions. For instance, digit constraints often arise in number problems.

Discover how to calculate the number of integers between 1 and 10^6 where the sum of their digits equals 9. This problem utilizes combinatorics, specifically the stars and bars theorem and binomial coefficients, making it suitable for Grades 10-12 students.

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