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Handshake Problem

Mediummathematics

At a party every pair of guests shakes hands exactly once, and the total number of handshakes recorded is 45; the number of guests present at the party equals which value?

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About This Question

Subject
mathematics
Chapter
permutations and combinations
Topic
handshake problem
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillhandshake-problemcombinationsquadratic-equationpair-counting

Solution

Correct Answer:

The total number of handshakes among n people, each pair shaking once, is C(n, 2) = n(n - 1)/2, a direct combination count used in JEE Advanced. Setting this equal to 45 gives n(n - 1)/2 = 45, so n(n - 1) = 90. Seeking two consecutive integers whose product is 90, we find 10 × 9 = 90, so n = 10. The equation n^2 - n - 90 = 0 factors as (n - 10)(n + 9) = 0, and the positive root is n = 10. Option 9 gives only C(9,2) = 36 handshakes. Option 45 confuses the number of handshakes with the number of guests. Option 15 gives C(15,2) = 105, far too many. Hence there are 10 guests. Plausibility check: with 10 guests, the number of distinct pairs is C(10,2) = 45, exactly the recorded handshake count, confirming the solution to the combination equation. Lexicographic ranking is a disciplined counting exercise in which, position by position, one tallies every arrangement that would alphabetically precede the target word. The method scales to words with repeated letters by using multinomial counts for the blocks of fixed prefixes, and careful bookkeeping of remaining available letters at each stage is what guarantees the final rank is exact.

This medium difficulty mathematics question is from the chapter permutations and combinations, covering the topic of handshake problem. It appeared in the 2025 exam.

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