Maths Olympiad Prep

Library / Switzerland — Second Round /7 of 9

Switzerland — Second Round, 2022

Combinatorics Difficulty 6.1 National Olympiad Prove it Switzerland

Problem:

a school class of n≥2n \geq 2 children is taking several group pictures. For every group with at least one child, there is exactly one picture containing this specific group. The pictures are now hung up in different rooms in the school, such that every child appears in at most one photo per room.

a. Show that this is possible if the school has 2n−12^{n-1} rooms.

b. Show that this is not possible if the school has less than 2n−12^{n-1} rooms.

Solution

Solution:

a.
One way of doing this is to pair each photo with its complement and put the photo with all children in a room by itself. Indeed, this is the only way (see appendix).

b.
The simplest way of doing this is to provide a set of 2n−12^{n-1} photos, of which no two are disjoint. There are many ways of doing so (taking all the photos containing a given child is one example, amongst many others) and these all provide the desired proof, as all these photos have to be in different rooms.

Otherwise, a more general counting argument also suffices: as at most nn kids appear per room, and there must be n2n−1n 2^{n-1} appearances in total, we need at least 2n−12^{n-1} rooms.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.