To solve the problem of how many 7-digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, and 4, we will follow these steps:
### Step 1: Identify the digits and their frequencies
We have the digits: 1, 2, 0, 2, 4, 2, and 4.
- The digit 2 appears 3 times.
- The digit 4 appears 2 times.
- The digit 1 appears 1 time.
- The digit 0 appears 1 time.
### Step 2: Calculate the total arrangements without restrictions
To find the total arrangements of these digits, we use 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 items (7 digits).
- \( n_1, n_2, \ldots \) are the frequencies of the repeating items.
In our case:
\[
\text{Total arrangements} = \frac{7!}{3! \times 2! \times 1! \times 1!}
\]
Calculating this:
\[
7! = 5040
\]
\[
3! = 6, \quad 2! = 2, \quad 1! = 1
\]
So,
\[
\text{Total arrangements} = \frac{5040}{6 \times 2 \times 1 \times 1} = \frac{5040}{12} = 420
\]
### Step 3: Exclude invalid arrangements (those starting with 0)
Next, we need to exclude the arrangements that start with 0, as they do not count as valid 7-digit numbers.
If 0 is the leading digit, we have the remaining digits: 1, 2, 2, 4, 2, 4 (6 digits).
Now we calculate the arrangements of these 6 digits:
\[
\text{Arrangements starting with 0} = \frac{6!}{3! \times 2! \times 1!}
\]
Calculating this:
\[
6! = 720
\]
So,
\[
\text{Arrangements starting with 0} = \frac{720}{6 \times 2 \times 1} = \frac{720}{12} = 60
\]
### Step 4: Calculate valid 7-digit numbers
Finally, we subtract the invalid arrangements from the total arrangements:
\[
\text{Valid 7-digit numbers} = \text{Total arrangements} - \text{Arrangements starting with 0}
\]
\[
\text{Valid 7-digit numbers} = 420 - 60 = 360
\]
Thus, the total number of valid 7-digit numbers that can be formed is **360**.
