How many different four digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 such that the digit 5 appears exactly once?

Options

A. 1024
B. 2048
C. 4096
D. none of these

Answer

To solve the problem of how many different four-digit numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 such that the digit 5 appears exactly once, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to form a four-digit number using the digits 1 to 9, ensuring that the digit '5' appears exactly once.

2. **Choosing the Position for '5'**: The digit '5' can occupy any one of the four positions in the four-digit number. Therefore, we have 4 choices for the position of '5'.

3. **Filling the Remaining Positions**: After placing '5', we need to fill the remaining three positions with digits from the set {1, 2, 3, 4, 6, 7, 8, 9} (note that '5' is excluded). This gives us 8 available digits.

4. **Calculating the Number of Combinations**:
- For each of the remaining three positions, we can choose any of the 8 digits.
- Therefore, the number of ways to fill these three positions is calculated as:
\[ 8 \times 8 \times 8 = 8^3 = 512 \]

5. **Total Combinations**: Since we have 4 different positions where '5' can be placed, we multiply the number of ways to fill the remaining positions by the number of choices for the position of '5':
\[ \text{Total combinations} = 4 \times 512 = 2048 \]

### Final Answer:
Thus, the total number of different four-digit numbers that can be formed under the given conditions is **2048**.

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