# There are 85 subsets with 5 numbers beetween {1,2,..,11}. proof there are 2 subsets (A and B) that min(A)=min(B) , med(A)=med(B), max(A)=max(B).

med (A) = 3rd num in the subset.

example : for A = {2,3,5,8,11} min (A) = 2 , med (A) = 5 , max (A) = 11.

I tried to solve it with Pigeonhole principle , but I can't find how to get 84 holes to 85 pigeons (= subsets).

• It would be better to use the term subset than subgroup in this context. Commented Sep 26, 2019 at 8:28
• Please make an effort to format your questions, see the excellent hints here Commented Sep 26, 2019 at 8:29
• Do you mean: prove that given any 85 subsets of $\{1,2,\dots,11\}$ with 5 elements, we can always find two with the same minimum, same median, and same maximum? Commented Sep 26, 2019 at 8:35
• yes @almagest , Commented Sep 26, 2019 at 8:38

If we take from each subset min, med and max element, they will form a triplet $$(a,b,c)$$ with the property that $$b>a+1$$, $$c>b+1$$. So we essentially are asking the question how many of these triplets exist.
We first place 8 empty boxes leaving gaps between them. There are 9 gaps if we count the space to the left or to the right as gap. There are $$\binom93=84$$ ways to place 3 boxes with balls in them. Thus, there are only 84 possible triplets of $$(\min,\mathrm{med}, \max)$$ that your subsets can have. Apply pigeonhole principle now.