To solve this question, we need to count how many integers greater than 2000 can be formed using the digits 0, 1, 2, 3, 4, and 5, with each digit being used at most once.

Step-by-Step Solution

1. Identify Possible Integer Lengths:

   • For an integer to be greater than 2000, it needs to be at least 4 digits long since the available digits only go up to 5.
   • So, we only need to consider 4-digit and 5-digit integers.

2. Determine the Range for 4-Digit Numbers (Starting with 2, 3, 4, or 5):

   • The leading (thousands) digit must be one of 2, 3, 4, or 5 to ensure the number is greater than 2000.

Case 1: 4-Digit Numbers

• Thousands Place: Choose from {2, 3, 4, 5} (4 choices).
• Remaining 3 Digits: Select from the remaining 5 digits (since one digit has been used in the thousands place).
• Permutations for Remaining Digits: Arrange the remaining 3 digits in 3! (6) ways.

So, the total number of 4-digit numbers is: $4 \times 5 \times 6 = 120$

Case 2: 5-Digit Numbers

• First Digit (Tens of Thousands Place): Choose from {2, 3, 4, 5} (4 choices), as before.
• Remaining 4 Digits: Choose 4 more digits from the 5 remaining ones.
• Permutations for Remaining Digits: Arrange the remaining 4 digits in 4! (24) ways.

So, the total number of 5-digit numbers is: $4 \times 24 = 96$

Final Calculation

Adding both cases together: $120 + 96 = 216$

Answer

The total number of integers greater than 2000 that can be formed is: $\boxed{216}$

Would you like further details or have any questions?

Related Questions:

1. How would the solution change if the number had to be greater than 3000?
2. How many numbers could be formed if repetition of digits were allowed?
3. What if the range was between 2000 and 4000? How would that affect the solution?
4. How many such numbers can be formed if we are limited to only 4-digit numbers?
5. What is the highest number that can be formed under these constraints?

Tip: For combinatorics problems, try to break down choices by positions and consider cases individually for clear counting.