On an island of 20 wizards, every set of three cast a spell on another. Show that there must be a wizard targeted by at least 9 wizards. > Twenty wizards meet on an island. Every set of 3 wizards cast a spell together on another wizard. Show that there must be a wizard who had been targeted by at least 9 wizards. (Each wizard can be in multiple groups.)

Let $d_i$ be the number of wizards that take action on wizard $W_i$. Then by double counting we have $${20\choose 3} =\sum {d_i\choose 3} \leq 20{d\choose 3}$$

where $d$ maximal $d_i$. If $d\leq 8$ we get $$ 19\cdot 18 \leq 8\cdot 7\cdot 6\implies 57\leq 56$$ A contradiction.