The number of integers greater than $2000$ that can be formed with the digits $0, 1, 2, 3, 4, 5$, using each digit at most once, is

We need to form integers greater than 2000 using the digits 0, 1, 2, 3, 4, 5, where each digit can be used at most once in each number.

Since we want numbers greater than 2000, we need to consider:

• 4-digit numbers starting with 2, 3, 4, or 5
• All 5-digit numbers (automatically greater than 2000)
• All 6-digit numbers (automatically greater than 2000)

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For a 4-digit number to be greater than 2000, it must start with 2, 3, 4, or 5.

First digit: 4 choices (2, 3, 4, or 5)

Second digit: 5 choices (any of the remaining 5 digits)

Third digit: 4 choices (any of the remaining 4 digits)

Fourth digit: 3 choices (any of the remaining 3 digits)

Total 4-digit numbers: $4 \times 5 \times 4 \times 3$

$= 240$

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Any 5-digit number is automatically greater than 2000, so we count all possible 5-digit numbers.

The first digit cannot be 0 (otherwise it wouldn't be a 5-digit number).

First digit: 5 choices (1, 2, 3, 4, or 5) - cannot use 0

Second digit: 5 choices (any of the remaining 5 digits, including 0)

Third digit: 4 choices (any of the remaining 4 digits)

Fourth digit: 3 choices (any of the remaining 3 digits)

Fifth digit: 2 choices (any of the remaining 2 digits)

Total 5-digit numbers: $5 \times 5 \times 4 \times 3 \times 2$

$= 600$

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Any 6-digit number is automatically greater than 2000, so we count all possible 6-digit numbers.

The first digit cannot be 0.

First digit: 5 choices (1, 2, 3, 4, or 5) - cannot use 0

Remaining 5 positions: We arrange the remaining 5 digits in $5!$ ways

Total 6-digit numbers: $5 \times 5!$

$= 5 \times 120$

$= 600$

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Total numbers greater than 2000: $240 + 600 + 600$

$= 1440$

When we solve permutation problems, we always identify the constraints first (like "greater than 2000" or "cannot start with 0"), then count systematically by position.