How many 5 digit numbers can be formed having the digits 0 three times and 3 two times?

Options

A. 4
B. 6
C. 8
D. 10

Answer

To solve the problem of how many 5-digit numbers can be formed using the digits 0 three times and 3 two times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to form a 5-digit number using the digits 0 (three times) and 3 (two times). A valid 5-digit number cannot start with 0.

2. **Identify the Positions**: Since we need a 5-digit number, we can represent the number as five boxes: _ _ _ _ _.

3. **Determine the First Digit**: The first digit cannot be 0, so it must be 3. We place 3 in the first box:
- 3 _ _ _ _

4. **Fill Remaining Boxes**: Now we have to fill the remaining four boxes with the digits 0 (three times) and 3 (one time, since we already used one 3):
- Remaining digits to place: 0, 0, 0, 3

5. **Calculate Arrangements**: We need to find the number of distinct arrangements of the digits 0, 0, 0, and 3 in the remaining four boxes. The formula for arrangements of n items where there are repetitions is given by:
\[ \text{Number of arrangements} = \frac{n!}{n_1! \times n_2!} \]
where \( n \) is the total number of items, and \( n_1, n_2, \ldots \) are the counts of each distinct item.

Here, we have:
- Total items (n) = 4 (three 0's and one 3)
- Repetitions: three 0's and one 3.

Thus, the number of distinct arrangements is:
\[ \text{Number of arrangements} = \frac{4!}{3! \times 1!} \]

6. **Calculate Factorials**:
- \( 4! = 24 \)
- \( 3! = 6 \)
- \( 1! = 1 \)

Plugging these values into the formula gives:
\[ \text{Number of arrangements} = \frac{24}{6 \times 1} = 4 \]

7. **Final Answer**: Therefore, the total number of 5-digit numbers that can be formed is **4**.