I saw this question but I couldn't find the answer. Assume that we have a 10x10 table, and it's filled with 0 to 9 numbers ( 10 of each of them are in the table, 10x zero, 10x one, and ... )
By using graph, prove that we have a column or a row that we can find at least 4 different numbers in it.
edit: I have found this method to create a graph from any table-matrix : https://www.math3ma.com/blog/matrices-probability-graphs
after that, I thought that it would be a good way to use coloured edges. and now the thing to prove is to have a vertex with at least four colour edges connected to it. maybe with pigeonhole, the problem can be solved but I couldn't
For each number $i$ from $0$ to $9$ let $r_i$ and $c_i$ be the number of rows and columns, respectively, containing the number $i$. Since all ten instances of $i$ are contained in a table with dimensions $r_i$ and $c_i$, we have $r_i\times c_i\ge 10$. It easily follows $r_i+c_i\ge 7$. It follows $\sum_i r_i+c_i\ge 70$. Wthout loss of generality we can suppose that $\sum r_i\ge 35$. The pigeonhole principle implies that there exists a row which is counted in at least four $r_i$.