Club Membership Audit and a Sharing Relation
At a higher secondary school, 88 students belong to at least one of three clubs: Chess (C), Quiz (Q) and Debate (D). The membership register gives |C| = 45, |Q| = 40, |D| = 35, |C ∩ Q| = 15, |Q ∩ D| = 12, |C ∩ D| = 10 and |C ∩ Q ∩ D| = 5. The secretary wants a one-page summary for the principal. She first writes 'students in exactly one club = 45 + 40 + 35 = 120', then notices the total exceeds 88. She also proposes a relation R on the 88 students by x R y whenever x and y share at least one club. To test the idea, she lists four students: Anbu (Chess, Quiz), Bala (Quiz, Debate), Chitra (Debate only) and Divya (Chess only). The secretary claims R is an equivalence relation because 'sharing a club is like being in the same group'. The principal needs defensible numbers, not a corrected guess.
1. Compute the number of students in exactly one club and the number in at least two clubs. Show how the given counts are combined, and explain the error in 45 + 40 + 35.
Read the worked answer
Use inclusion–exclusion first to check the totals.
|C ∪ Q ∪ D| = 45 + 40 + 35 - 15 - 12 - 10 + 5 = 88, which matches the register.
For exactly one club, start with 45 + 40 + 35 = 120. A student in exactly two clubs has been counted twice, so subtract two extra copies of each pairwise overlap: subtract 2(15 + 12 + 10). A student in all three clubs was counted three times and would then have been subtracted six times, giving a contribution of -3; add 3 × 5 to bring that contribution to 0. Hence
exactly one = 120 - 2(37) + 15 = 120 - 74 + 15 = 61.
Exactly two = (15 + 12 + 10) - 3(5) = 37 - 15 = 22.
At least two = exactly two + all three = 22 + 5 = 27.
Check: 61 + 22 + 5 = 88. The error in 120 is that it counts exactly-two students twice and exactly-three students three times; it is not a count of exactly one.
2. Test the secretary's claim that R is an equivalence relation using Anbu, Bala, Chitra and Divya. If it fails, identify the property and give the smallest change to the definition that would repair it.
Read the worked answer
Reflexive: every listed student belongs to at least one club, so each shares a club with themselves; R is reflexive. Symmetric: if x shares a club with y, then y shares that same club with x; R is symmetric.
Transitive: Anbu R Bala because both are in Quiz. Bala R Chitra because both are in Debate. But Anbu and Chitra share no club, so Anbu is not related to Chitra. Therefore R is not transitive and cannot be an equivalence relation.
A repair is to define x R y only when x and y belong to exactly the same set of clubs. That relation is reflexive, symmetric and transitive, so it is an equivalence relation. The original 'share at least one club' relation is only reflexive and symmetric.