Geometry pigeonhole principle problem.

Question

let sets: $A_1 , A_2 , A_3 , ..., A_{13} \subset [10]$

$\forall i : |A_i|=6$

I'm asked to show that there exist $1\le j_1 \lt j_2 \lt j_3 \le 13$,

such that: $|A_{j_1}\cap A_{j_2}\cap A_{j_3}| \ge 3 $

all i got so far is that if i know how many sequences of size 6 out of $[10]$, which is $ {10 \choose 6 } =210$, and if i know how many sequences have the same three numbers, then i could use the pigeonhole principle and say that there must be 3 groups that has 3 numbers in common, but i'm missing the second part. any help would be greatly appreciated, more of all i wanna completely understand on how to solve it.

Answer 1

Every $A_i$ contains ${6\choose 3}$ 3-subsets. Therefore, in total, the $A_i$'s combined contain $13{6\choose 3}=260$ 3-subsets.

We now count the number of possible different 3-subsets. In total, there are ${10\choose3}=120$ different possible 3-subsets. Thus, there are 260 3-subsets across the $A_i$'s, of which 120 are different. Now by the pigeonhole principle, there must be a 3-subset that appears 3 times across the $A_i$'s.