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?
- (a) 6
- (b) 8
- (c) 10 ✓ UPSC's answer
- (d) 12
Why the answer is (c)
• The PIN is a 3-digit decreasing sequence chosen from digits 1 to 7, so once three digits are selected, their left-to-right order is fixed.
• The condition that any two digits differ by at least 2 means no two selected digits can be consecutive.
• The possible decreasing PINs are 531, 631, 731, 641, 741, 751, 642, 742, 752, and 753.
• There are 10 such PINs, and each corresponds to exactly one valid 3-digit set.
• Therefore the maximum number of attempts needed to try all possible PINs with certainty is 10, matching option (c).
Why the other options are wrong
- (a) 6
- Option (a) 6 is too low because the valid PINs include 10 distinct decreasing sequences, such as 531, 631, 731, 641, 741, 751, 642, 742, 752, and 753.
- (b) 8
- Option (b) 8 is too low because the no-consecutive condition leaves 10 valid 3-digit sets from 1 to 7, not 8.
- (d) 12
- Option (d) 12 is too high because only 10 3-digit subsets of {1, 2, 3, 4, 5, 6, 7} have pairwise differences of at least 2, so 12 attempts exceed the total possible PINs.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.