CSAT Solved Papers/ 2021/Q69

2021 CSAT — Q69

Quant Counting & combinatorics 2.5 marks Medium

There are 66 persons arranged in a row. Another person has to shake hands with 33 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place?

  1. A 3
  2. B 4 Answer
  3. C 5
  4. D 6

Worked rationale

Choose 33 of the 66 persons in a row with no two chosen persons adjacent. The standard count for choosing kk non-adjacent items from nn in a line is

(n−k+1k)=(6−3+13)=(43)=4.\binom{n-k+1}{k} = \binom{6-3+1}{3} = \binom{4}{3} = 4.

Listing confirms it (positions 11–66, no two consecutive): {1,3,5},{1,3,6},{1,4,6},{2,4,6}\{1,3,5\}, \{1,3,6\}, \{1,4,6\}, \{2,4,6\} — exactly 44.

Answer: (b) 4.

Why the other options miss

  • A
    a valid triple missed: overlooks one legal choice, e.g. forgets {1,3,6}\{1,3,6\} or {1,4,6}\{1,4,6\}.
  • C
    let an adjacent pair through: includes one triple containing a consecutive pair (e.g. {1,3,4}\{1,3,4\} slips in).
  • D
    the formula misapplied: uses (n−k+1k)\binom{n-k+1}{k} wrongly, or counts ordered arrangements instead of selections.

Specialist insight

The clean tool is the gap (stars-and-bars) formula (n−k+1k)\binom{n-k+1}{k} for kk non-adjacent picks from a line of nn — here (43)=4\binom{4}{3} = 4. If you’d rather list, the discipline is to start each triple with the smallest legal first element and slide the others rightward, which both guarantees no adjacency and prevents double-counting. Either way, 44, and the off-by-one distractor (c) is precisely the “let one adjacent pair sneak in” error.

The trap, in one line

Non-adjacent picks: (6−3+13)=(43)=4⇒\binom{6-3+1}{3} = \binom{4}{3} = 4 \Rightarrow (b).

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