Number of five digit integers, with sum of the digits equal to 43 are :

Options

A. 5
B. 10
C. 15
D. 35

Answer

To solve the problem of finding the number of five-digit integers where the sum of the digits equals 43, we will analyze the possible combinations of digits that can achieve this sum.

### Step-by-Step Solution:

1. **Understanding the Constraints**:
- Each digit in a five-digit number can range from 0 to 9.
- The maximum sum of five digits (if all were 9) is \(9 + 9 + 9 + 9 + 9 = 45\).
- Therefore, to achieve a sum of 43, we need to have some digits that are less than 9.

2. **Identifying Possible Cases**:
- The only way to achieve a sum of 43 with five digits is to have four digits equal to 9 and one digit less than 9. This leads us to two cases:
- **Case 1**: Four 9s and one 7.
- **Case 2**: Three 9s and two 8s.

3. **Calculating for Case 1**:
- In this case, the digits are \(9, 9, 9, 9, 7\).
- To find the number of distinct arrangements of these digits, we use the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{5!}{4!} = 5 \]
- This is because we have 5 total digits, with one digit (7) appearing once and the digit 9 appearing four times.

4. **Calculating for Case 2**:
- In this case, the digits are \(9, 9, 9, 8, 8\).
- Again, we use the permutations formula:
\[ \text{Number of arrangements} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10 \]
- Here, we have 5 total digits, with the digit 9 appearing three times and the digit 8 appearing two times.

5. **Total Count of Five-Digit Integers**:
- Now, we sum the results from both cases:
\[ \text{Total} = 5 + 10 = 15 \]

### Final Answer:
The total number of five-digit integers with the sum of the digits equal to 43 is **15**.