The number of 3 digit odd number, when repetition of digits is allowed is

Options

A. 450
B. 360
C. 400
D. 420

Answer

To find the number of 3-digit odd numbers when repetition of digits is allowed, we can break down the problem step by step.

### Step 1: Identify the structure of a 3-digit odd number
A 3-digit number can be represented as XYZ, where:
- X is the hundreds place
- Y is the tens place
- Z is the units place (which must be odd for the number to be odd)

### Step 2: Determine the choices for the units place (Z)
Since the number must be odd, Z can only be one of the odd digits. The odd digits are:
- 1, 3, 5, 7, 9

This gives us a total of **5 choices** for Z.

### Step 3: Determine the choices for the tens place (Y)
For the tens place, we can use any digit from 0 to 9 (since repetition is allowed). This gives us a total of **10 choices** for Y.

### Step 4: Determine the choices for the hundreds place (X)
For the hundreds place, we cannot use 0 (as it would make it a 2-digit number). Therefore, the valid digits for X are:
- 1, 2, 3, 4, 5, 6, 7, 8, 9

This gives us a total of **9 choices** for X.

### Step 5: Calculate the total number of 3-digit odd numbers
Now, we can multiply the number of choices for each digit:
- Choices for X (hundreds place) = 9
- Choices for Y (tens place) = 10
- Choices for Z (units place) = 5

Total number of 3-digit odd numbers = Choices for X × Choices for Y × Choices for Z
= 9 × 10 × 5 = 450

### Final Answer
The total number of 3-digit odd numbers when repetition of digits is allowed is **450**.

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