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

Explanation

To solve the problem of finding the number of seven-digit integers formed by using the digits 1, 2, and 3, with the sum of the digits equal to 10, we can follow these steps:

### Step 1: Identify the possible cases
We need to find combinations of the digits 1, 2, and 3 that add up to 10 while using exactly 7 digits.

#### Case 1:
Using five 1's and two 2's:
- The sum will be: \(1 + 1 + 1 + 1 + 1 + 2 + 2 = 10\)

#### Case 2:
Using four 1's and three 2's:
- The sum will be: \(1 + 1 + 1 + 1 + 2 + 2 + 2 = 10\)

These are the only two cases that satisfy the conditions.

### Step 2: Calculate permutations for each case
Now we will calculate the number of permutations for each case.

#### For Case 1 (five 1's and two 2's):
- The arrangement can be calculated using the formula for permutations of multiset:
\[ \text{Number of arrangements} = \frac{7!}{5! \cdot 2!} \]
Calculating this:
\[ = \frac{7 \times 6}{2 \times 1} = 21 \]

#### For Case 2 (four 1's and three 2's):
- The arrangement can be calculated similarly:
\[ \text{Number of arrangements} = \frac{7!}{4! \cdot 3!} \]
Calculating this:
\[ = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 3: Add the results from both cases
Now, we add the number of arrangements from both cases to get the total number of seven-digit integers:
\[ \text{Total} = 21 + 35 = 56 \]

### Final Answer
Thus, the total number of seven-digit integers formed by using the digits 1, 2, and 3, with the sum of the digits equal to 10, is **56**.

---