How many 3 digit number can be formed whose unit digit is always zero and repetition of digits is not allowed ?

Answer

To solve the problem of how many three-digit numbers can be formed where the unit digit is always 0 and repetition of digits is not allowed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of the Three-Digit Number**:
A three-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, \(C\) must always be 0.

2. **Determine the Choices for the Hundreds Place (A)**:
Since the unit digit is fixed as 0, the hundreds place \(A\) can be filled with any digit from 1 to 9 (as 0 cannot be in the hundreds place). Therefore, there are 9 possible choices for \(A\).

3. **Determine the Choices for the Tens Place (B)**:
After choosing a digit for \(A\), we cannot use that digit again (since repetition is not allowed). Additionally, we cannot use 0 (since it is already in the units place). Thus, for the tens place \(B\), we have 8 remaining choices (the digits 1-9 excluding the digit chosen for \(A\)).

4. **Calculate the Total Number of Combinations**:
The total number of three-digit numbers can be calculated by multiplying the number of choices for \(A\) and \(B\):
\[ \text{Total Combinations} = \text{Choices for } A \times \text{Choices for } B = 9 \times 8 = 72 \]

### Final Answer:
Therefore, the total number of three-digit numbers that can be formed under the given conditions is **72**.

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