CAT 2022 Slot 2 Paper

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. $5 \times 4 \times 3 = 60 \text{ ways}$ Total number of possible 4-digit numbers $= 4 \times 60 = 240$ Case 2: 5-digit numbers First digit cannot be zero. Therefore, total number of cases $= 5 \times 5 \times 4 \times 3 \times 2 = 600$ Case 3: 6-digit numbers First digit cannot be zero. Therefore, total number of cases $= 5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600$ Total number of integers possible $= 600 + 600 + 240 = 1440$ The answer is option A.