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?

Key conditions: choose 3 distinct digits from 1 to 7, digits are already fixed in decreasing order, and any two chosen digits differ by at least 2. So we only need to count valid 3-digit sets.

• Starting with 1: possible sets are {1,3,5}, {1,3,6}, {1,3,7}, {1,4,6}, {1,4,7}, {1,5,7} → 6 valid PINs.
• Starting with 2: possible sets are {2,4,6}, {2,4,7}, {2,5,7} → 3 valid PINs.
• Starting with 3: only {3,5,7} → 1 valid PIN.
• Starting with 4 or more: impossible to choose 3 digits with gap at least 2.

Total valid PINs = 6 + 3 + 1 = 10. In the worst case, one may need to try all valid PINs to be certain.

Option-wise verdict:

• (a) 6: Incorrect — valid PINs are more than 6.
• (b) 8: Incorrect — count reaches 10.
• (c) 10: Correct — total valid PINs = 10.
• (d) 12: Incorrect — only 10 satisfy all conditions.