# Proving minimum exsistence of intersection cardinality

Let $F_1,F_2...F_{13}$ be sets such that $\forall 1\le i \le 13: F_i\subseteq [10]$ and $|F_i|=6$ when $[10]={1,2,3...10}$

prove that there are $1 \le j < k < l \le 13$ such that $|F_j \cap F_k \cap F_l| \ge 3$

It's pretty obvious I should use the Pigeonhole principle, but cluless to how should I get started?

The way you can set this up is to make a table in which rows correspond to the $F_i$'s and the columns correspond to 3-element subsets of $[10]$. Then you can say that you will mark with a + a cell such that the 3-element set corresponding to that cell's column is a subset of the $F_i$ corresponding to that cell's row.
$$\begin{array}{c|c c c c} & \{0,1,2\} & \{0,1,3\} & \cdots & \{8,9,10\}\\ \hline F_1&-&+&\cdots&-\\ F_2&+&+&\cdots&-\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ F_{13}&-&-&\cdots&+ \end{array}$$