Find max N people with 2022 different colored balls choosing two balls each so that no 3 people exactly contain balls of 3 colours

Question

Here is the "exact" wording given in the problem.

There are $N$ people. Each person gets two balls of different colours among $2022$ balls with different colours. The combinations of the ball colours they get are different from each other. Find the maximum possible value of $N$ such that no $3$ people exactly contain balls of $3$ colours.

The English sounds a little strange to me as I am still questioning whether I understood the problem correctly. The given "Answer key" (no solution) states $1022121$.

The only number coincidence I can see is that $(2022)(2022)/4 = (1011)^2 = (1011)(1011) = 1022121$ which could mean nothing.

Here are some thoughts regarding how or why I got stuck:

1. Number of ways two get two balls of different colours among $2022$ balls with different colours is $(2022)(2021)/2 = 2043231$.

2. If we choose three colours say $a, b, c$ then the pairs containing those colours are $(a, b), (b,c)$ or $(c, a)$ where order is not important, i.e. there are "three" of them.

3. Number of different three colour combinations which can be derived from $2022$ different colours is $C(2022, 3) = 1 375, 775, 540$.

4. Help part

How do I link these to count the answer of $1022121$? Am I missing something or did I misunderstand the question? I have relatively average level secondary school level Math Olympiad abilities.

Any help is appreciated. Thank you.

Jon

Answer 1

We consider the set $A= \left \{ 1, 2, ..., 2022 \right \}$ and $2$ subset $A_1, A_2$. We consider the ordered pair $(x,y)$ wher $x \in A_1, y \in A_2$. We call $A_1, A_2$ good if there aren't $3$ pair of the form $(a,b),(b,c),(a,c)$ created as explained above. We will prove that if $A_1, A_2$ are good then $k \in A_1 \Leftrightarrow k\notin A_2$. If it is true, then $max(|A_1|+|A_2|)=2022$. Now suppose by contradiction that $g\in A_1,A_2$, and they are good so we take $a_1 \in A_1, a_2 \in A_2$, the we can meke the order pairs $(a_1,g),(g,a_2),(a_1,a_2)$, so the contradiction. So we have proved that $k \in A_1 \Leftrightarrow k\notin A_2$. The number of ordere pairs is $|A_1|×|A_2|$. So for maximize this product with costant sum $2022$, we take $|A_1|=1011,|A_2|=1011$. We can see that it work, so we are done. Sorry for my bad english.