
Problem 3. Let $A$ be a set with 8 elements. Find the maximal number of 3 -element subsets of $A$, such that the intersection of any two of them is not a 2-element set.


Solution. Let $B_{1}, B_{2}, \ldots, B_{n}$ be subsets of $A$ such that $\left|B_{i}\right|=3,\left|B_{i} \cap B_{j}\right| \neq 2(i, j=1, \ldots, n)$. Assume that there exists an element $a \in A$, which belongs to four of the subsets $B_{1}, B_{2}, \ldots, B_{n}$ (e.g. $\left.a \in B_{1}, B_{2}, B_{3}, B_{4}\right)$. Then $\left|B_{i} \cap B_{j}\right| \geq 1(i, j=1, \ldots, 4)$. But $B_{i} \neq B_{j}$ if $i \neq j$, i.e. $\left|B_{i} \cap B_{j}\right| \neq 3$. Thus $\left|B_{i} \cap B_{j}\right|=1(i, j=1,2,3,4)$. It follows from here that $|A| \geq 1+4.2=9$,
which is a contradiction. Therefore every element of $A$ belongs to at most three of the subsets $B_{1}, B_{2}, \ldots, B_{n}$. Then $3 n \leq 8.3$, i.e. $n \leq 8$.

If $A=\left\{a_{1}, a_{2}, \ldots, a_{8}\right\}$, then the subsets

$$
\begin{array}{llll}
B_{1}=\left\{a_{1}, a_{2}, a_{3}\right\}, & B_{2}=\left\{a_{1}, a_{4}, a_{5}\right\}, & B_{3}=\left\{a_{1}, a_{6}, a_{7}\right\}, & B_{4}=\left\{a_{8}, a_{3}, a_{4}\right\}, \\
B_{5}=\left\{a_{8}, a_{2}, a_{6}\right\}, & B_{6}=\left\{a_{8}, a_{5}, a_{7}\right\}, & B_{7}=\left\{a_{3}, a_{5}, a_{6}\right\}, & B_{8}=\left\{a_{2}, a_{4}, a_{7}\right\}
\end{array}
$$

provide an example of exactly eight 3 -element subsets of $A$, such that $\left|B_{i} \cap B_{j}\right| \neq 2$.

Therefore the searched number is $n=8$.

## Grade 10 .
