Forming Integers Greater Than 2000

Digits available: {0, 1, 2, 3, 4, 5} (total 6 digits). We need numbers > 2000 using digits at most once.

Case 1: 4-digit numbers > 2000 First digit can be 2, 3, 4, 5 (4 choices). Remaining 3 places chosen from remaining 5 digits: $^5P_3 = 5 \times 4 \times 3 = 60$. Total 4-digit numbers = $4 \times 60 = 240$.

Case 2: 5-digit numbers First digit cannot be 0 (5 choices). Remaining 4 places: $^5P_4 = 120$. Total 5-digit numbers = $5 \times 120 = 600$.

Case 3: 6-digit numbers First digit cannot be 0 (5 choices). Remaining 5 places: $^5P_5 = 120$. Total 6-digit numbers = $5 \times 120 = 600$.

Total integers = $240 + 600 + 600 = 1440$.