XXI OBM

Let $A_i\subset \{1,2,\dots ,10\}$ be the set of the positions of the tokens in the $i$$Theequalityholdsifandonlyif5ofthe$k_i${}^{\prime }sequal4andtheother5equal3.Inthiscase,$\sum_{1 \le i \le 10} \binom{k_i}{2} = \binom{10}{2}$andhenceeachsubsetof$\{1, 2, \dots, 10\}$with2elementsshouldbeinexactlyone$A_i$.<br/><br/>Thereforeifitwerepossibletoconstructaninstancewith35tokens,eachelementof$\{1, 2, \dots, 10\}$wouldeitherbein3subsetswith4elementsorin1subsetwith4elementsand3subsetswith3elements.Sincethereare5subsetswith4elements,theremustbeelementswhichbelongto3subsetswith4elements.Wemaythussupposewlogthat$A_1 = \{1, 2, 3, 4\}$,$A_2 = \{1, 5, 6, 7\}$,$A_3 = \{1, 8, 9, 10\}$.Howeveranyothersubsetwith4elementswouldbecontainedin$\{2, 3, \dots, 10\}$andthereforeitsintersectionwithoneof$A_1$,$A_2$or$A_3$wouldhaveatleast2elements.Weconcludethatitisimpossibletohave$\sum_{1\le i\le 10} k_i = 35$.

On the other hand, there exist instances with 34 tokens: