To find the number of integers between 1 and 1000 whose sum of digits equals 12, we can break down the problem step by step.
### Step 1: Understand the problem
We need to find integers in the range from 1 to 1000. The integers can be represented as three-digit numbers, where we can include leading zeros (e.g., 001, 002, ..., 999). Therefore, we are looking for non-negative integers \( x_1, x_2, x_3 \) such that:
\[ x_1 + x_2 + x_3 = 12 \]
where \( x_1, x_2, x_3 \) represent the digits of the number.
### Step 2: Set up the equation
Since each digit \( x_i \) (for \( i = 1, 2, 3 \)) must be between 0 and 9, we can use the stars and bars combinatorial method. However, we first need to consider the restriction that each \( x_i \) must be less than or equal to 9.
### Step 3: Calculate total combinations without restrictions
Using the stars and bars theorem, the total number of non-negative integer solutions to the equation \( x_1 + x_2 + x_3 = 12 \) is given by:
\[
\text{Total solutions} = \binom{n + r - 1}{r - 1}
\]
where \( n = 12 \) and \( r = 3 \):
\[
\text{Total solutions} = \binom{12 + 3 - 1}{3 - 1} = \binom{14}{2}
\]
Calculating \( \binom{14}{2} \):
\[
\binom{14}{2} = \frac{14 \times 13}{2} = 91
\]
### Step 4: Subtract invalid combinations
Next, we need to subtract the cases where one or more digits exceed 9.
#### Case 1: One digit exceeds 9
Assume \( x_1 > 9 \). Let \( x_1' = x_1 - 10 \), then we have:
\[
x_1' + x_2 + x_3 = 2
\]
The number of non-negative integer solutions is:
\[
\binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6
\]
Since any of the three digits can exceed 9, we multiply by 3:
\[
3 \times 6 = 18
\]
### Step 5: Calculate valid combinations
Now, we subtract the invalid combinations from the total combinations:
\[
\text{Valid combinations} = 91 - 18 = 73
\]
### Final Answer
The number of integers between 1 and 1000 having their sum of digits equal to 12 is **73**.
---
