To solve the problem, we need to find the number of 7-digit numbers formed using the digits 1, 2, and 3 such that the sum of the digits equals 10. We will denote this count as \( K \).
### Step 1: Understanding the Problem
We need to find combinations of the digits 1, 2, and 3 that add up to 10 while ensuring that the total number of digits used is 7.
### Step 2: Setting Up the Equation
Let:
- \( x_1 \) = number of 1's
- \( x_2 \) = number of 2's
- \( x_3 \) = number of 3's
We have the following equations:
1. \( x_1 + x_2 + x_3 = 7 \) (total digits)
2. \( x_1 + 2x_2 + 3x_3 = 10 \) (sum of digits)
### Step 3: Solving the Equations
From equation 1, we can express \( x_3 \) in terms of \( x_1 \) and \( x_2 \):
\[ x_3 = 7 - x_1 - x_2 \]
Substituting \( x_3 \) into equation 2:
\[ x_1 + 2x_2 + 3(7 - x_1 - x_2) = 10 \]
\[ x_1 + 2x_2 + 21 - 3x_1 - 3x_2 = 10 \]
\[ -2x_1 - x_2 + 21 = 10 \]
\[ -2x_1 - x_2 = -11 \]
\[ 2x_1 + x_2 = 11 \]
Now we have a system of equations:
1. \( x_1 + x_2 + x_3 = 7 \)
2. \( 2x_1 + x_2 = 11 \)
### Step 4: Solving for \( x_1 \) and \( x_2 \)
From the second equation, we can express \( x_2 \):
\[ x_2 = 11 - 2x_1 \]
Substituting this into the first equation:
\[ x_1 + (11 - 2x_1) + x_3 = 7 \]
\[ -x_1 + 11 + x_3 = 7 \]
\[ x_3 = x_1 - 4 \]
### Step 5: Finding Non-negative Integer Solutions
Now we have:
1. \( x_1 + x_2 + (x_1 - 4) = 7 \)
2. \( x_2 = 11 - 2x_1 \)
We need \( x_1 \geq 4 \) (since \( x_3 \) must be non-negative). Let's consider possible values for \( x_1 \):
- If \( x_1 = 4 \), then \( x_2 = 3 \), \( x_3 = 0 \).
- If \( x_1 = 5 \), then \( x_2 = 1 \), \( x_3 = 1 \).
- If \( x_1 = 6 \), then \( x_2 = -1 \) (not valid).
So the valid combinations are:
1. \( (x_1, x_2, x_3) = (4, 3, 0) \)
2. \( (x_1, x_2, x_3) = (5, 1, 1) \)
### Step 6: Counting the Arrangements
Now we calculate the number of arrangements for each case.
**Case 1:** \( (4, 3, 0) \)
- The arrangement is \( \frac{7!}{4!3!} = \frac{5040}{24 \cdot 6} = 35 \).
**Case 2:** \( (5, 1, 1) \)
- The arrangement is \( \frac{7!}{5!1!1!} = \frac{5040}{120 \cdot 1 \cdot 1} = 42 \).
### Step 7: Total Count \( K \)
Adding both cases:
\[ K = 35 + 42 = 77 \]
### Step 8: Final Calculation
We need to find \( \frac{K + 46}{100} \):
\[ \frac{77 + 46}{100} = \frac{123}{100} = 1.23 \]
### Conclusion
The value of \( \frac{K + 46}{100} \) is \( 1.23 \).
