# Is it possible to pack the tetrominoes exactly on the checkboard?

I have this as an exercise but I'm not sure what's the best way to approach it. I've tried the pack them but wasn't able to do so. I see that having them combined will yield in 10 whites and 10 blacks ( because every piece will cover 2 blacks and 2 whites) which is exactly the chessboard but not sure if that's it.

every piece will cover 2 blacks and 2 whites

Are you sure about that? Have another look before looking at the spoiler below.

The T piece will cover 1 black and 3 white (or 3 black and 1 white). Every other piece covers 2 black and 2 white. So in total they will cover 9 black and 11 white (or 11 black and 9 white). Since the board have 10 black and 10 white, it is not possible.

This one is simple.

It is not possible to cover the 5x4 board with the given tetrominoes because

The total number of cells is $20$, $10$ black and $10$ white.

All cover $2$ black and $2$ white cells. Except one

That one, covers either 1 black and 3 white or 1 white and 3 black. Since, the counts comes to $8+1$ for black and $8+3$ for white (or vice-verse). We would never have all $10$ white and all $10$ black cells covered.