How many different eight digit numbers can be formed using only four digits 1,2,3,4 such that the digit 2 occurs twice?

Options

A. 20412
B. 12042
C. 25065
D. none of these

Answer

To find how many different eight-digit numbers can be formed using the digits 1, 2, 3, and 4, with the condition that the digit 2 occurs exactly twice, we can follow these steps:

### Step 1: Understand the composition of the number
We need to form an 8-digit number using the digits 1, 2, 3, and 4, where the digit 2 appears twice. This means we will have:
- 2 (occurs twice)
- 6 other digits which can be 1, 3, or 4.

### Step 2: Determine the total number of digits
The total number of digits in the number is 8. Since the digit 2 appears twice, we need to fill the remaining 6 positions with the digits 1, 3, and 4.

### Step 3: Set up the distribution of the remaining digits
Let’s denote the number of times we use the digits 1, 3, and 4 as \( x_1, x_3, \) and \( x_4 \) respectively. We need to satisfy the equation:
\[ x_1 + x_3 + x_4 = 6 \]
where \( x_1, x_3, \) and \( x_4 \) are non-negative integers.

### Step 4: Use the stars and bars method
The problem of distributing 6 indistinguishable items (the remaining digits) into 3 distinguishable boxes (the digits 1, 3, and 4) can be solved using the stars and bars theorem. The number of ways to do this is given by:
\[ \text{Number of solutions} = \binom{n+k-1}{k-1} \]
where \( n \) is the number of items to distribute (6) and \( k \) is the number of boxes (3).

Thus, we have:
\[ \text{Number of solutions} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28 \]

### Step 5: Calculate the arrangements of the digits
Now, we need to arrange the 8 digits: 2, 2, and the 6 digits which are either 1, 3, or 4. The total arrangements of these digits can be calculated using the formula for permutations of a multiset:
\[ \text{Arrangements} = \frac{n!}{n_1! \cdot n_2! \cdots n_k!} \]
where \( n \) is the total number of items, and \( n_1, n_2, \ldots, n_k \) are the counts of each distinct item.

In our case, we have:
- Total digits = 8
- Two 2's, and the remaining 6 digits can be any combination of 1's, 3's, and 4's.

Thus, the number of arrangements is:
\[ \text{Arrangements} = \frac{8!}{2! \cdot x_1! \cdot x_3! \cdot x_4!} \]

### Step 6: Combine the results
Since we have 28 ways to choose how many 1's, 3's, and 4's to use, and for each choice, we can arrange the digits in \( \frac{8!}{2! \cdot x_1! \cdot x_3! \cdot x_4!} \) ways, the total number of different 8-digit numbers is:
\[ \text{Total Numbers} = 28 \cdot \frac{8!}{2! \cdot x_1! \cdot x_3! \cdot x_4!} \]

### Final Calculation
Calculating \( 8! = 40320 \) and \( 2! = 2 \):
\[ \text{Total Numbers} = 28 \cdot \frac{40320}{2} = 28 \cdot 20160 = 564480 \]

Thus, the total number of different eight-digit numbers that can be formed is **564480**.