Towers 'n Tets problem.

Question

There are $n$ students in a school In this school there exists $n+1$ 3-member recreational clubs with no two clubs having the same 3 members. Show there exists two clubs with exactly 1 member in common

i have reading solution but i don't understand solution :Now, you might wonder why the problem has that name. Indeed, suppose the contrary. One way to visualize the three-sets is to create an $n\times(n+1)$ grid, and shade in $3$ boxes per column. We can show that the only "closed" (any un-closed subset is less efficient in a way we describe later) configurations that satisfy the conditions are $\{a,b,x_,1\}; \{a,b,x_2\},\dots$ (Tower based on the shape of the Schliesser Diagram) and $\{a,b,c\};\{a,b,d\};\{a,c,d\};\{b,c,d\}$ (Tetrahedron based on the shape of the Schliesser Diagram). But the first configuration has a column-row efficiency of $-2$, and the second has one of $0$, so there is no combination of them that can total to an efficiency of $1$, that of the grid

Answer 1

The proof is by contradiction: we assume that no two clubs share a single member, and show that in that case, the number of clubs is at most the number of students.

Pick a club with members $\{a,b,c\}$ and consider all other clubs that $a$, $b$, or $c$ are in. There are two cases:

1. Some pair $a$ and $b$ are in two other clubs together: there are also clubs $\{a,b,d\}$ and $\{a,b,e\}$. Now no other club can include $a$ and $c$ together: to avoid sharing only one member with $\{a,b,d\}$ it would also need $d$, but then it shares only one member with $\{a,b,e\}$. Similarly, no other club can include $b$ and $c$ together. So we are left with some number of clubs $$\{a,b,c\}, \{a,b,d\}, \{a,b,e\}, \{a,b,f\}, \{a,b,g\}, \dots$$
2. Each pair of people in $\{a,b,c\}$ is in at most one other club together. There can be clubs $\{a,b,x\}$, $\{a,c,y\}$, and $\{b,c,z\}$, but because they cannot share exactly one member, we must have $x=y=z$. This gives us the four clubs $$\{a,b,c\}, \{a,b,d\}, \{a,c,d\}, \{b,c,d\}$$ (though not all four clubs actually have to exist).

In both cases, any other club that used any of those same people would have to share a member with $\{a,b,c\}$, so it would already be in that collection of clubs. So in general, the students at the school can be split into some number of groups, with different groups not having any students in common, and each group forming clubs like case 1 or 2 above.

In each group, the number of clubs is at most the number of students (in case 1, there are actually two fewer clubs than students; in case 2, the number of clubs is the same as the number of students). So when all groups are combined, the number of clubs is at most $n$.

If the number of clubs were $n+1$, the assumption that no two clubs share a single member would have to be violated.

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There is also a more general result that if $n$ people are in clubs of odd size with no two clubs sharing an odd number of people, then there can be at most $n$ clubs. It has a proof using linear algebra.