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 can break it down step by step.

### Step 1: Understand the Constraints
A five-digit integer can have digits ranging from 0 to 9. However, since we are looking for a five-digit integer, the first digit cannot be 0. The maximum sum of five digits, each being 9, is \(9 + 9 + 9 + 9 + 9 = 45\). Therefore, a sum of 43 is achievable.

### Step 2: Identify Possible Combinations
We need to find combinations of five digits \(a, b, c, d, e\) such that:
\[ a + b + c + d + e = 43 \]

Given that the maximum digit is 9, we can explore the possible cases:

#### Case 1: Four digits are 9, and one digit is 7
- This gives us the combination: \(9, 9, 9, 9, 7\).
- The number of ways to arrange these digits is given by the formula for permutations of multiset:
\[ \text{Ways} = \frac{5!}{4!} = 5 \]

#### Case 2: Three digits are 9, two digits are 8
- This gives us the combination: \(9, 9, 9, 8, 8\).
- The number of ways to arrange these digits is:
\[ \text{Ways} = \frac{5!}{3! \times 2!} = \frac{120}{6 \times 2} = 10 \]

### Step 3: Calculate Total Combinations
Now, we sum the number of ways from both cases:
\[ \text{Total Ways} = 5 + 10 = 15 \]

### Step 4: Conclusion
Thus, the total number of five-digit integers with a sum of digits equal to 43 is **15**.