While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place?
Answer & explanation
Answer: (c) 85
Fix the hundreds digit h (7, 8 or 9). Any two different digits smaller than h, written in decreasing order, give exactly one valid number, so the count is C(7, 2) + C(8, 2) + C(9, 2) = 21 + 28 + 36 = 85.
- Hundreds digit 7: tens and units are two different digits from 0–6, written in decreasing order: C(7, 2) = 21.
- Hundreds digit 8: two digits from 0–7: C(8, 2) = 28.
- Hundreds digit 9: two digits from 0–8: C(9, 2) = 36.
- 1000 does not qualify. Total = 21 + 28 + 36 = 85.
- Check for 7: tens digit 6 allows 6 numbers (760–765), tens 5 allows 5, and so on: 6 + 5 + 4 + 3 + 2 + 1 = 21.
Remember · For strictly decreasing digits, choosing the digits fixes their order: count with combinations, not arrangements.
Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·