Interesting Sets

Since $|A_1 \cap A_i| = 1$ for all $1 \le i \le 2018$ and $|A_1| = 44$ , by the Pigeonhole Principle , there exists an element $a \in A_1$ such that 46 other sets also contain $a$ . Without loss of generality, let $a \in A_1, A_2, ..., A_{47}$ .

Suppose there exists a set $A_q$ such that $a \notin A_q$ . Then $A_q \cap A_1 \ne A_q \cap A_2 \ne ... \ne A_q \cap A_{47}$ , otherwise it would break the condition that $|A_j \cap A_k| = 1$ for integers $1 \le j < k \le 2018$ . However, this also means that $|A_q| \ge 47$ , which is a contradiction. Hence no such $A_q$ exists.

This also means that $\displaystyle \bigcap _{ i=1 }^{ 2018 }{ A_{ i } } =a$ . Hence

$\displaystyle \large \left| \bigcup _{ i=1 }^{ 2018 }{ A_{ i } } \right| = 2018 \times 43 + 1 = 86775$

The second condition seems interesting. huh?? $1\le j<k\le 2018$ and $|A_j\cap A_k|=1.$ ?? well. Let us imagine that there is an element say $X$ such that every set contains $X$ . And all other elements are distinct. So the second condition is ensured and the first condition helps us to count the answer.

$A_1$ has 44 distinct elements. $A_2$ has 43 elements which are not in A_1. $A_3$ also has 43 elements which are not in $A_1$ or $A_2$ ........ and so on........ $A_{2018}$ has 43 elements which are not in $A_1,A_2,A_3,.....,A_{2017}$

thus the answer is $44+2017*43$ = $86775$