▁and ▁the ▁ones ▁from ▁the ▁last ▁two ▁categories ▁count ▁twice . ▁This ▁results ▁in ▁ 8 4 ▁× ▁ 8 ▁+ ▁( 9 + 7 + 4 ) ▁× ▁ 4 ▁+ ▁( 3 + 1 ) ▁× ▁ 2 ▁= ▁ 7 6 0 ▁fixed ▁he pt om ino es . ▁ ▁Pack ing ▁and ▁t iling ▁Of ▁the ▁ 1 0 8 ▁free ▁he pt om ino es , ▁ 1 0 1 ▁satisfy ▁the ▁Con way ▁criter ion ▁and ▁ 3 ▁more ▁can ▁form ▁a ▁patch ▁satisfying ▁the ▁criter ion . ▁Thus , ▁only ▁ 4 ▁he pt om ino es ▁fail ▁to ▁satisfy ▁the ▁criter ion ▁and , ▁in ▁fact , ▁these ▁ 4 ▁are ▁unable ▁to ▁t ess ell ate ▁the ▁plane . ▁ ▁Although ▁a ▁complete ▁set ▁of ▁the ▁ 1 0 8 ▁free ▁he pt om ino es ▁has ▁a ▁total ▁of ▁ 7 5 6 ▁squares , ▁it ▁is ▁not ▁possible ▁to ▁tile ▁a ▁rectangle ▁with ▁that ▁set . ▁The ▁proof ▁of ▁this ▁is ▁trivial , ▁since ▁there ▁is ▁one ▁he pt om ino ▁which ▁has ▁a ▁hole . ▁It ▁is ▁also ▁impossible ▁to ▁pack ▁them ▁into ▁a ▁ 7 5 7 - square ▁rectangle ▁with ▁a ▁one - square ▁hole ▁because ▁ 7 5 7 ▁is ▁a ▁prime ▁number . ▁ ▁However , ▁the ▁set ▁of ▁ 1 0 7 ▁simply ▁connected ▁free ▁he pt om ino es — that ▁is , ▁the ▁ones ▁without ▁the ▁hole — can ▁tile ▁a ▁ 7 ▁by