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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
Case 1: 4-digit numbers
Given digits - 0, 1, 2, 3, 4, 5
_, _, _, _
As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.
For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways
Total number of possible 4-digit numbers = 4*60 = 240
Case 2: 5-digit numbers
_, _, _, _, _
First digit cannot be zero.
Therefore, total number of cases = 5*5*4*3*2 = 600
Case 3: 6-digit numbers
_, _, _, _, _, _
First digit cannot be zero.
Therefore, total number of cases = 5*5*4*3*2*1 = 600
Total number of integers possible = 600 + 600 + 240 = 1440
The answer is option A.
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