The number of 3- digit numbers containing the digit 7 exactly once :

Options

A. 225
B. 220
C. 200
D. 180

Answer

To find the number of 3-digit numbers containing the digit 7 exactly once, we can break the problem into three cases based on the position of the digit 7 in the 3-digit number.

### Step 1: Case 1 - 7 at the Last Position
- If 7 is at the last position, the number can be represented as **_ _ 7**.
- The first digit (hundreds place) cannot be 0 (since it is a 3-digit number), and it also cannot be 7 (since we want 7 to appear exactly once).
- Therefore, the first digit can be any digit from 1 to 9 except 7, which gives us 8 choices (1, 2, 3, 4, 5, 6, 8, 9).
- The second digit (tens place) can be any digit from 0 to 9 except 7, giving us 9 choices (0, 1, 2, 3, 4, 5, 6, 8, 9).
- Thus, the total number of combinations for this case is:
\[ \text{Choices for first digit} \times \text{Choices for second digit} = 8 \times 9 = 72 \]

### Step 2: Case 2 - 7 in the Middle Position
- If 7 is in the middle, the number can be represented as **_ 7 _**.
- The first digit (hundreds place) again cannot be 0 or 7, giving us 8 choices (1, 2, 3, 4, 5, 6, 8, 9).
- The last digit (units place) can be any digit from 0 to 9 except 7, giving us 9 choices.
- Thus, the total number of combinations for this case is:
\[ \text{Choices for first digit} \times \text{Choices for last digit} = 8 \times 9 = 72 \]

### Step 3: Case 3 - 7 at the First Position
- If 7 is at the first position, the number can be represented as **7 _ _**.
- The second digit (tens place) can be any digit from 0 to 9 except 7, giving us 9 choices.
- The last digit (units place) can also be any digit from 0 to 9 except 7, giving us another 9 choices.
- Thus, the total number of combinations for this case is:
\[ \text{Choices for second digit} \times \text{Choices for last digit} = 9 \times 9 = 81 \]

### Step 4: Total Count
Now, we add the results from all three cases to get the total number of 3-digit numbers containing the digit 7 exactly once:
\[ \text{Total} = 72 + 72 + 81 = 225 \]

### Final Answer
The total number of 3-digit numbers containing the digit 7 exactly once is **225**.

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