The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

Options

A. 55
B. 66
C. 77
D. 88

Answer

To solve the problem of finding the number of seven-digit integers formed by the digits 1, 2, and 3, with the sum of the digits equal to 10, we can break it down into cases based on the possible combinations of these digits.

### Step 1: Identify the Cases
We need to find combinations of the digits 1, 2, and 3 that add up to 10. The digits can be used multiple times, but we need to ensure that we have a total of 7 digits.

#### Case 1: 5 Ones, 1 Two, and 1 Three
- Here, we have 5 digits of 1, 1 digit of 2, and 1 digit of 3.
- The total sum is: \(5 \times 1 + 1 \times 2 + 1 \times 3 = 5 + 2 + 3 = 10\).

#### Case 2: 4 Ones and 3 Twos
- Here, we have 4 digits of 1 and 3 digits of 2.
- The total sum is: \(4 \times 1 + 3 \times 2 = 4 + 6 = 10\).

### Step 2: Calculate the Number of Arrangements for Each Case

#### For Case 1: 5 Ones, 1 Two, and 1 Three
- The total number of digits is 7.
- The number of arrangements can be calculated using the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{7!}{5! \cdot 1! \cdot 1!} \]
Calculating this gives:
\[ = \frac{5040}{120 \cdot 1 \cdot 1} = \frac{5040}{120} = 42 \]

#### For Case 2: 4 Ones and 3 Twos
- The total number of digits is again 7.
- The number of arrangements is:
\[ \text{Number of arrangements} = \frac{7!}{4! \cdot 3!} \]
Calculating this gives:
\[ = \frac{5040}{24 \cdot 6} = \frac{5040}{144} = 35 \]

### Step 3: Total Number of Seven-Digit Integers
Now, we add the results from both cases to find the total number of seven-digit integers:
\[ \text{Total} = 42 + 35 = 77 \]

Thus, the number of seven-digit integers formed by the digits 1, 2, and 3 with a sum of 10 is **77**.

### Final Answer
The final answer is **77**.