CSAT Solved Papers/ 2022/Q55

2022 CSAT — Q55

Quant Counting & combinatorics 2.5 marks Hard

There is a numeric lock which has a 33-digit PIN. The PIN contains digits 11 to 77. 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 22. How many maximum attempts does one need to find out the PIN with certainty?

  1. A 6
  2. B 8
  3. C 10 Answer
  4. D 12

Worked rationale

The number of attempts to be certain equals the number of valid PINs. A PIN is a strictly decreasing triple a>b>ca > b > c from {1,…,7}\{1,\dots,7\} with adjacent differences ≥2\ge 2: a−b≥2a - b \ge 2 and b−c≥2b - c \ge 2 (these force every pair to differ by ≥2\ge 2).

Collapse the gaps by a shift: let p=c, q=b−1, r=a−2p = c,\ q = b - 1,\ r = a - 2. Then b≥c+2⇔q≥p+1b \ge c + 2 \Leftrightarrow q \ge p+1 and a≥b+2⇔r≥q+1a \ge b + 2 \Leftrightarrow r \ge q+1, so p<q<rp < q < r, with p≥1p \ge 1 and r=a−2≤5r = a - 2 \le 5.

Thus the valid PINs correspond bijectively to strictly increasing triples p<q<rp < q < r in {1,…,5}\{1,\dots,5\}:

(53)=10.\binom{5}{3} = 10.

Answer: (c) 10.

Why the other options miss

  • A
    only the tightest triples: counts just the triples with both gaps exactly 22 (e.g. 7 5 37\,5\,3), missing the ones with wider gaps.
  • B
    wrong reduced range: uses (53)\binom{5}{3} on a mis-sized shifted set (range 44 instead of 55).
  • D
    order counted or gap relaxed: counts ordered triples, or loosens the gap to ”≥1\ge 1 somewhere,” over-counting.

Specialist insight

The clean tool is the gap-removing bijection: subtracting 0,1,20,1,2 from the three positions turns the “differ by ≥2\ge 2” constraint into a plain strictly-increasing choice, so the count is just (53)=10\binom{5}{3} = 10. Decreasing-order plus a minimum gap is exactly a (n−(k−1)(g−1)k)\binom{n - (k-1)(g-1)}{k} pattern; here n=7,k=3,g=2n=7, k=3, g=2 gives (7−23)=(53)=10\binom{7-2}{3} = \binom{5}{3} = 10.

The trap, in one line

Min-gap-22 decreasing triples from 11–77 map to (53)=10\binom{5}{3} = 10 plain triples ⇒\Rightarrow (c).

← All 2022 CSAT questions