UPSC Prelims 2023 CSAT Paper II · Q47 of 77 Mental Ability medium

What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated ?

  1. (a) 7998 ✓ UPSC's answer
  2. (b) 8028
  3. (c) 8878
  4. (d) 9238

Why the answer is (a)

• The 4-digit numbers must be less than 2000, so the thousands digit must be 1.

• The remaining three digits (2, 3, 4) can be arranged in the hundreds, tens, and units places in 3! = 6 ways.

• The sum of the digits in the thousands place is 1 × 6 = 6.

• Each of the digits 2, 3, and 4 appears in the hundreds, tens, and units places exactly 2! = 2 times each.

• The sum of the digits 2, 3, and 4 is 9; thus, the sum for each of the remaining places is 9 × 2 = 18.

• The total sum is calculated as (6 × 1000) + (18 × 100) + (18 × 10) + (18 × 1) = 6000 + 1800 + 180 + 18 = 7998.

Why the other options are wrong

(b) 8028
Option 8028 is incorrect because the total sum of the valid permutations is 7998, not 8028.
(c) 8878
Option 8878 is incorrect because it does not match the calculated sum of 7998 derived from the positional values.
(d) 9238
Option 9238 is incorrect because it exceeds the actual sum of 7998 obtained by summing all valid 4-digit numbers.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2023, held on 28 May 2023. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.

Reading the answer is not the same as getting it right under a clock. Practise this question with UPSC's negative marking, and anything you miss goes into an error notebook until you get it right twice.

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