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 find the number of seven-digit integers formed using the digits 1, 2, and 3, such that the sum of the digits equals 10, we can break the problem down into cases based on the combinations of these digits.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find seven-digit integers using the digits 1, 2, and 3, where the sum of the digits equals 10. The digits can be repeated, and we need to ensure that the total number of digits is 7.

2. **Setting Up the Equation**:
Let \( x_1 \), \( x_2 \), and \( x_3 \) be the number of times the digits 1, 2, and 3 are used, respectively. We have the following conditions:
- \( x_1 + x_2 + x_3 = 7 \) (total digits)
- \( x_1 + 2x_2 + 3x_3 = 10 \) (sum of digits)

3. **Solving the Equations**:
From the first equation, we can express \( x_3 \) in terms of \( x_1 \) and \( x_2 \):
\[ x_3 = 7 - x_1 - x_2 \]
Substituting this into the second equation gives:
\[ x_1 + 2x_2 + 3(7 - x_1 - x_2) = 10 \]
Simplifying this:
\[ x_1 + 2x_2 + 21 - 3x_1 - 3x_2 = 10 \]
\[ -2x_1 - x_2 + 21 = 10 \]
\[ -2x_1 - x_2 = -11 \quad \Rightarrow \quad 2x_1 + x_2 = 11 \]

4. **Finding Possible Values**:
Now we have two equations:
- \( x_1 + x_2 + x_3 = 7 \)
- \( 2x_1 + x_2 = 11 \)

We can express \( x_2 \) in terms of \( x_1 \):
\[ x_2 = 11 - 2x_1 \]
Substituting this into the first equation:
\[ x_1 + (11 - 2x_1) + x_3 = 7 \]
\[ -x_1 + x_3 + 11 = 7 \]
\[ x_3 = x_1 - 4 \]

5. **Finding Non-Negative Solutions**:
We need \( x_1 \), \( x_2 \), and \( x_3 \) to be non-negative:
- From \( x_3 = x_1 - 4 \), we get \( x_1 \geq 4 \).
- From \( x_2 = 11 - 2x_1 \), we need \( 11 - 2x_1 \geq 0 \) which gives \( x_1 \leq 5.5 \). Since \( x_1 \) must be an integer, \( x_1 \) can be 4 or 5.

**Case 1**: \( x_1 = 4 \)
- Then \( x_2 = 11 - 2(4) = 3 \)
- \( x_3 = 4 - 4 = 0 \)

**Case 2**: \( x_1 = 5 \)
- Then \( x_2 = 11 - 2(5) = 1 \)
- \( x_3 = 5 - 4 = 1 \)

6. **Calculating Combinations**:
For each case, we calculate the number of arrangements:
- **Case 1**: (4 ones, 3 twos, 0 threes)
\[ \text{Number of arrangements} = \frac{7!}{4!3!} = \frac{5040}{24 \times 6} = 35 \]

- **Case 2**: (5 ones, 1 two, 1 three)
\[ \text{Number of arrangements} = \frac{7!}{5!1!1!} = \frac{5040}{120 \times 1 \times 1} = 42 \]

7. **Total Combinations**:
Adding the results from both cases gives:
\[ \text{Total} = 35 + 42 = 77 \]

### Final Answer:
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 **77**.