There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
Answer & explanation
Answer: (c) 10
Each set of three digits from 1–7 that are pairwise at least 2 apart gives exactly one PIN (written in decreasing order). There are 10 such sets, so in the worst case 10 attempts are needed.
- Once three digits are chosen, the decreasing order fixes the PIN, so count the digit sets.
- The digits must be from 1–7 with every two at least 2 apart. List by the largest digit: 753, 752, 751, 742, 741, 731, 642, 641, 631, 531 — 10 PINs.
- Quick count: choosing 3 of 7 numbers with no two consecutive = C(7 − 3 + 1, 3) = C(5, 3) = 10.
- In the worst case the right PIN is the last one tried, so up to 10 attempts are needed.
Remember · Choosing k of n numbers with no two consecutive: C(n − k + 1, k).
Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·