 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-4/6e/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

For finitely many finite sets

A_{1}, \dots, A_{k}

i = 1 ⋃ k A_{i} = \emptyset \neq = J \subseteq {1, \dots, k} \sum (- 1)^{∣ J ∣ + 1} j \in J ⋂ A_{j} .

To see why, an element belonging to exactly

t \geq 1

of the sets contributes

\sum_{j = 1}^{t} (- 1)^{j + 1} (j t) = 1

by the binomial theorem. Elements outside the union contribute zero. Equivalently, if these sets lie in a finite universe

U

, the number avoiding all of them is

∣ U ∣ + \sum_{\emptyset \neq = J} (- 1)^{∣ J ∣} ⋂_{j \in J} A_{j} .

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