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?
Answer & explanation
Answer: (a) 7998
Numbers below 2000 must start with 1, and the other three places take 2, 3 and 4 in 3! = 6 orders. The thousands digits give 6 × 1000 = 6000; in each other place each of 2, 3 and 4 appears twice, adding 18 × 111 = 1998. The total is 7998.
- Below 2000 with the digits 1, 2, 3, 4 used once each, the first digit must be 1; 2, 3, 4 fill the rest in 3! = 6 ways.
- Thousands place: 6 × 1000 = 6000.
- In each of the hundreds, tens and units places, each of 2, 3, 4 appears 2 times: (2 + 3 + 4) × 2 = 18.
- Their contribution = 18 × (100 + 10 + 1) = 18 × 111 = 1998.
- Total = 6000 + 1998 = 7998.
Remember · Sum of numbers formed by permutations: count how often each digit sits in each place, then multiply by the place values (111…).
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·