To solve the problem of how many different numbers greater than 1000 can be formed using the digits 1, 0, 2, and 3, we will follow these steps:
### Step 1: Identify the conditions
We need to form numbers greater than 1000 using the digits 1, 0, 2, and 3. Since the number must be greater than 1000, the first digit (thousands place) cannot be 0.
### Step 2: Determine the valid first digits
The valid digits for the first position (thousands place) are 1, 2, or 3. This gives us 3 options for the first digit.
### Step 3: Count the remaining digits
After choosing the first digit, we have 3 remaining digits to choose from for the other positions. The digits available will depend on which digit we chose first.
### Step 4: Calculate the number of arrangements
For each choice of the first digit, we can arrange the remaining 3 digits in the other three positions (hundreds, tens, and units). The number of arrangements of 3 digits is given by 3! (3 factorial), which is equal to 6.
### Step 5: Total combinations
Now, we multiply the number of choices for the first digit by the number of arrangements of the remaining digits:
- Number of choices for the first digit = 3 (1, 2, or 3)
- Number of arrangements of the remaining digits = 6 (3!)
Thus, the total number of different numbers greater than 1000 is:
\[
3 \times 6 = 18
\]
### Conclusion
The statement that "Only 24 numbers with different digits greater than 1000 can be formed from the digits 1, 0, 2, 3" is incorrect. The correct number of such combinations is 18.
