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. $5\times 4\times 3=60\mathrm{ways}$
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.