How many 7-digit numbers can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

Explanation

To solve the problem of how many 7-digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4, we will follow these steps:

### Step 1: Count the total arrangements without restrictions
We start by calculating the total arrangements of the digits. The total number of digits is 7, where:
- The digit '2' appears 3 times.
- The digit '4' appears 2 times.
- The digit '1' and '0' appear 1 time each.

The formula for permutations of a multiset is given by:

\[ \text{Total arrangements} = \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \]

Where \( n \) is the total number of items, and \( n_1, n_2, \ldots, n_k \) are the frequencies of the distinct items.

Here, we have:

\[ n = 7, \quad n_1 = 3 \text{ (for '2')}, \quad n_2 = 2 \text{ (for '4')}, \quad n_3 = 1 \text{ (for '1')}, \quad n_4 = 1 \text{ (for '0')} \]

Thus, the total arrangements are:

\[ \text{Total arrangements} = \frac{7!}{3! \times 2! \times 1! \times 1!} = \frac{5040}{6 \times 2 \times 1 \times 1} = \frac{5040}{12} = 420 \]

### Step 2: Exclude arrangements starting with '0'
Since we are forming a 7-digit number, the first digit cannot be '0'. We need to calculate how many arrangements start with '0' and subtract these from the total arrangements.

If '0' is the first digit, we are left with the digits 1, 2, 2, 2, 4, 4. The number of arrangements of these 6 digits is:

\[ \text{Arrangements with '0' as first digit} = \frac{6!}{3! \times 2! \times 1!} = \frac{720}{6 \times 2 \times 1} = \frac{720}{12} = 60 \]

### Step 3: Calculate valid arrangements
Now, we subtract the arrangements that start with '0' from the total arrangements:

\[ \text{Valid arrangements} = \text{Total arrangements} - \text{Arrangements with '0' as first digit} = 420 - 60 = 360 \]

### Final Answer
The total number of valid 7-digit numbers that can be formed using the digits 1, 2, 0, 2, 4, 2, 4 is **360**.

---