Solution-
First 9 letters of alphabet are- A,B,C,D,E,F,G,H,I
Total number of ways of selection of 4 letters from 9 alphabets = 9C4
=9!÷((4!).(9-4)!) = 9!÷(4!×5!) = (9×8×7×6)÷(24) = 126
The number of ways of arranging these 4 numbers = 4! = 24
∴ Total number of possible permutations = 9C4×4! = 126×24 = 3024
∴ option number 2 is correct.
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