How many 6-digit numbers can be formed by using the digits 4, 5, 0, 3, 4, 5?

Explanation

To solve the problem of how many 6-digit numbers can be formed using the digits 4, 5, 0, 3, 4, and 5, we will follow these steps:

### Step 1: Identify the digits and their counts
The digits given are 4, 5, 0, 3, 4, 5. We can summarize the counts of each digit:
- 4 appears 2 times
- 5 appears 2 times
- 0 appears 1 time
- 3 appears 1 time

### Step 2: Calculate the total arrangements without restrictions
The total number of arrangements of the digits can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \]
Where \( n \) is the total number of digits, and \( n_1, n_2, \ldots \) are the counts of each distinct digit.

Here, \( n = 6 \) (total digits), \( n_1 = 2 \) (for digit 4), \( n_2 = 2 \) (for digit 5), \( n_3 = 1 \) (for digit 0), and \( n_4 = 1 \) (for digit 3).

Thus, the total arrangements are:
\[ \text{Total arrangements} = \frac{6!}{2! \times 2! \times 1! \times 1!} = \frac{720}{2 \times 2 \times 1 \times 1} = \frac{720}{4} = 180 \]

### Step 3: Exclude arrangements starting with 0
Since a 6-digit number cannot start with 0, we need to subtract the arrangements that start with 0.

If 0 is the first digit, the remaining digits are 4, 5, 3, 4, 5. The counts for these digits are:
- 4 appears 2 times
- 5 appears 2 times
- 3 appears 1 time

The arrangements of these 5 digits are:
\[ \text{Arrangements starting with 0} = \frac{5!}{2! \times 2! \times 1!} = \frac{120}{2 \times 2 \times 1} = \frac{120}{4} = 30 \]

### Step 4: Calculate valid 6-digit numbers
Now, we subtract the invalid arrangements (those starting with 0) from the total arrangements:
\[ \text{Valid 6-digit numbers} = \text{Total arrangements} - \text{Arrangements starting with 0} = 180 - 30 = 150 \]

### Final Answer
Thus, the total number of valid 6-digit numbers that can be formed is **150**.

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