Number of 3 digit numbers that can be formed having unit digit as zero and repetition of digit is allowed is

Options

A. 72
B. 81
C. 100
D. 90

Answer

To solve the problem of finding the number of 3-digit numbers that can be formed with the unit digit as zero and allowing repetition of digits, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure of the 3-Digit Number**:
A 3-digit number can be represented as \(abc\), where \(a\) is the hundreds place, \(b\) is the tens place, and \(c\) is the units place. According to the problem, the unit digit \(c\) must be 0.

2. **Fixing the Unit Digit**:
Since the unit digit \(c\) is fixed as 0, we can write the number as \(ab0\). Now we need to determine the values for \(a\) and \(b\).

3. **Choosing the Hundreds Place (Digit \(a\))**:
The hundreds place \(a\) cannot be 0 (as it would not be a 3-digit number), so \(a\) can take any digit from 1 to 9. This gives us a total of 9 options for \(a\).

4. **Choosing the Tens Place (Digit \(b\))**:
The tens place \(b\) can be any digit from 0 to 9, since repetition of digits is allowed. This gives us a total of 10 options for \(b\).

5. **Calculating the Total Number of Combinations**:
Since the choices for \(a\) and \(b\) are independent, we can multiply the number of options for each digit to find the total number of 3-digit numbers:
\[ \text{Total combinations} = (\text{choices for } a) \times (\text{choices for } b) = 9 \times 10 = 90 \]

### Final Answer:
The total number of 3-digit numbers that can be formed with the unit digit as 0 and allowing repetition of digits is **90**.

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