To solve the problem of finding the number of eight-digit integers formed using the digits 1, 2, and 3, such that the sum of the digits equals 12, we can break it down into a systematic approach. Here’s a step-by-step solution:
### Step 1: Define the Variables
Let:
- \( x_1 \) = number of 1's
- \( x_2 \) = number of 2's
- \( x_3 \) = number of 3's
We need to satisfy the following conditions:
1. \( x_1 + x_2 + x_3 = 8 \) (since we are forming an 8-digit number)
2. \( x_1 + 2x_2 + 3x_3 = 12 \) (since the sum of the digits must equal 12)
### Step 2: Solve the Equations
From the first equation, we can express \( x_3 \) in terms of \( x_1 \) and \( x_2 \):
\[
x_3 = 8 - x_1 - x_2
\]
Substituting \( x_3 \) into the second equation gives:
\[
x_1 + 2x_2 + 3(8 - x_1 - x_2) = 12
\]
Simplifying this:
\[
x_1 + 2x_2 + 24 - 3x_1 - 3x_2 = 12
\]
\[
-2x_1 - x_2 + 24 = 12
\]
\[
-2x_1 - x_2 = -12
\]
\[
2x_1 + x_2 = 12
\]
### Step 3: Find Possible Values
Now we have two equations:
1. \( x_1 + x_2 + x_3 = 8 \)
2. \( 2x_1 + x_2 = 12 \)
From the second equation, we can express \( x_2 \):
\[
x_2 = 12 - 2x_1
\]
Substituting this into the first equation:
\[
x_1 + (12 - 2x_1) + x_3 = 8
\]
\[
- x_1 + 12 + x_3 = 8
\]
\[
x_3 = x_1 - 4
\]
### Step 4: Determine Non-Negative Integer Solutions
Now we need to ensure that \( x_1 \), \( x_2 \), and \( x_3 \) are non-negative integers:
1. From \( x_3 = x_1 - 4 \), we require \( x_1 \geq 4 \).
2. From \( x_2 = 12 - 2x_1 \), we require \( 12 - 2x_1 \geq 0 \) or \( x_1 \leq 6 \).
Thus, \( x_1 \) can take values 4, 5, or 6.
### Step 5: Calculate Cases
- **Case 1**: \( x_1 = 4 \)
- \( x_2 = 12 - 2(4) = 4 \)
- \( x_3 = 4 - 4 = 0 \)
The arrangement is \( 4 \) ones, \( 4 \) twos, and \( 0 \) threes:
\[
\text{Permutations} = \frac{8!}{4!4!} = 70
\]
- **Case 2**: \( x_1 = 5 \)
- \( x_2 = 12 - 2(5) = 2 \)
- \( x_3 = 5 - 4 = 1 \)
The arrangement is \( 5 \) ones, \( 2 \) twos, and \( 1 \) three:
\[
\text{Permutations} = \frac{8!}{5!2!1!} = 336
\]
- **Case 3**: \( x_1 = 6 \)
- \( x_2 = 12 - 2(6) = 0 \)
- \( x_3 = 6 - 4 = 2 \)
The arrangement is \( 6 \) ones, \( 0 \) twos, and \( 2 \) threes:
\[
\text{Permutations} = \frac{8!}{6!0!2!} = 28
\]
### Step 6: Total the Cases
Now, we sum the permutations from all cases:
\[
70 + 336 + 28 = 434
\]
### Final Answer
The total number of eight-digit integers formed using the digits 1, 2, and 3 such that their sum equals 12 is **434**.
