How many three-digit numbers can be formed by using the digits 1, 3, 6 and 8 when the digits may be repeated any no. of times?

Options

A. 48
B. 64
C. 80
D. 32

Answer

To solve the problem of how many three-digit numbers can be formed using the digits 1, 3, 6, and 8 with repetition allowed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Digits Available**:
We have the digits 1, 3, 6, and 8 available for forming the three-digit numbers.

2. **Determine the Number of Choices for Each Digit**:
Since repetition is allowed, we can use any of the 4 digits (1, 3, 6, 8) for each position in the three-digit number.

3. **Calculate the Choices for Each Position**:
- For the **hundreds place**, we have 4 choices (1, 3, 6, or 8).
- For the **tens place**, we again have 4 choices (1, 3, 6, or 8).
- For the **units place**, we also have 4 choices (1, 3, 6, or 8).

4. **Multiply the Choices Together**:
Since the choices for each digit are independent, we multiply the number of choices for each position:
\[ \text{Total Numbers} = \text{Choices for Hundreds} \times \text{Choices for Tens} \times \text{Choices for Units} \]
\[ \text{Total Numbers} = 4 \times 4 \times 4 \]

5. **Calculate the Final Result**:
\[ \text{Total Numbers} = 4^3 = 64 \]

Thus, the total number of three-digit numbers that can be formed is **64**.