When rotations and reflections are not considered to be distinct shapes, there are 35 different free hexominoes.
[4] Although a complete set of 35 hexominoes has a total of 210 squares, it is not possible to pack them into a rectangle.
A simple way to demonstrate that such a packing of hexominoes is not possible is via a parity argument.
[5] It is also possible for two sets of pieces to fit a rectangle of size 420, or for the set of 60 one-sided hexominoes (18 of which cover an even number of black squares) to fit a rectangle of size 360.
A polyhedral net for the cube cannot contain the O-tetromino, nor the I-pentomino, the U-pentomino, or the V-pentomino.