Problem

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Tags: pigeonhole principle, combinatorics unsolved, combinatorics



All members of the senate were firstly divided into $S$ senate commissions . According to the rules, no commission has less that $5$ senators and every two commissions have different number of senators. After the first session the commissions were closed and new commissions were opened. Some of the senators now are not a part of any commission. It resulted also that every two senators that were in the same commission in the first session , are not any more in the same commission. (a)Prove that at least $4S+10$ senators were left outside the commissions. (b)Prove that this number is achievable. Albanian National Mathematical Olympiad 2010---12 GRADE Question 5.