The number of distinct positive integers can be formed using $0,1,2,3$ where each integer used at most once…

The number of distinct positive integers can be formed using $0,1,2,3$ where each integer used at most once is equal to
  1. $84$
  2. $64$
  3. $48$
  4. $36$

Solution

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

Asked in: AP EAMCET 2021 (23 Aug Shift 2)

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