How many different four digit numbers can be formed using digits 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition such that the digit 5 appears exactly once?

Options

A. 1243
B. 1234
C. 1344
D. 1355

Answer

To solve the problem of how many different four-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition such that the digit 5 appears exactly once, we will follow these steps:

### Step 1: Identify the positions for digit 5
The digit 5 can occupy any of the four positions in the four-digit number. Therefore, we have four cases to consider:
1. 5 is in the first position.
2. 5 is in the second position.
3. 5 is in the third position.
4. 5 is in the fourth position.

### Step 2: Calculate the arrangements for each case
For each case, we will fill the remaining three positions with the digits from the set {1, 2, 3, 4, 6, 7, 8, 9} (since 5 is already used). There are 8 digits left to choose from.

#### Case 1: 5 in the first position
- First position: 5
- Remaining positions: 3 positions to fill with digits from {1, 2, 3, 4, 6, 7, 8, 9}.
- We can choose any digit for the second position (8 options), any remaining digit for the third position (7 options), and any remaining digit for the fourth position (6 options).

The number of arrangements for this case is:
\[ 8 \times 7 \times 6 = 336 \]

#### Case 2: 5 in the second position
- First position: 8 options (any digit except 5)
- Remaining positions: 7 options for the third position and 6 options for the fourth position.

The number of arrangements for this case is:
\[ 8 \times 7 \times 6 = 336 \]

#### Case 3: 5 in the third position
- First position: 8 options (any digit except 5)
- Second position: 7 options (any digit except 5 and the digit chosen for the first position).
- Fourth position: 6 options.

The number of arrangements for this case is:
\[ 8 \times 7 \times 6 = 336 \]

#### Case 4: 5 in the fourth position
- First position: 8 options (any digit except 5)
- Second position: 7 options (any digit except 5 and the digit chosen for the first position).
- Third position: 6 options.

The number of arrangements for this case is:
\[ 8 \times 7 \times 6 = 336 \]

### Step 3: Total arrangements
Now, we sum the arrangements from all four cases:
\[ 336 + 336 + 336 + 336 = 1344 \]

Thus, the total number of different four-digit numbers that can be formed using the digits 1 to 9, with the digit 5 appearing exactly once, is **1344**.

### Final Answer
The total number of different four-digit numbers is **1344**.

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