The number of integers between 1 and 1000 having their sum of digits equal to 12 is

Options

A. 405
B. 410
C. 415
D. 420

Answer

To solve the problem of finding the number of integers between 1 and 1000 whose sum of digits equals 12, we can follow these steps:

### Step 1: Define the digits
We need to consider three digits \(X_1\), \(X_2\), and \(X_3\) that represent the hundreds, tens, and units places of a three-digit number (or less). The condition is that \(X_1 + X_2 + X_3 = 12\).

### Step 2: Set the bounds for the digits
Each digit \(X_i\) (where \(i = 1, 2, 3\)) must satisfy \(0 \leq X_i \leq 9\). This means that we need to find the non-negative integer solutions to the equation \(X_1 + X_2 + X_3 = 12\) under these constraints.

### Step 3: Use the stars and bars method
Without any constraints, the number of non-negative integer solutions to the equation \(X_1 + X_2 + X_3 = 12\) can be calculated using the stars and bars theorem:
\[ \text{Number of solutions} = \binom{n + r - 1}{r - 1} = \binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} \]

### Step 4: Calculate \(\binom{14}{2}\)
Calculating \(\binom{14}{2}\):
\[ \binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91 \]

### Step 5: Subtract cases where any digit exceeds 9
Now we need to subtract the cases where any of the digits exceeds 9. If, for example, \(X_1 > 9\), we can set \(X_1' = X_1 - 10\) (where \(X_1' \geq 0\)). The equation becomes:
\[ X_1' + X_2 + X_3 = 2 \]
The number of non-negative integer solutions to this equation is:
\[ \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6 \]
This same calculation holds for \(X_2\) and \(X_3\) exceeding 9, so we subtract 6 for each case.

### Step 6: Total cases to subtract
Since there are three digits, we subtract a total of:
\[ 3 \times 6 = 18 \]

### Step 7: Calculate the final result
Now we can find the total number of valid combinations:
\[ \text{Total valid combinations} = 91 - 18 = 73 \]

### Conclusion
Thus, the number of integers between 1 and 1000 whose sum of digits equals 12 is **73**.

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