AP EAPCET 23-Aug-2021 Shift 2 Paper

Given integers, 0,1,2,3 Case I The number of ways to form four digit integers =3×3×2×1=18
(first place can be filled by one of the three choices 1,2,3)
Case II The number of 3 -digit integers =3×3×2=18
Case III The number of 2 -digit integers =3×3=9
Case IV The number of 1 -digit integer =3
∴ The number of distinct positive integers =18+18+9+3=48