Tiling a $2n-1 \times 2n-1$ board by $L$ triominos, $Z$ tetrominoes and box tetrominoes

Question

A $2n-1 \times 2n-1$ board is going to be tiled by L triominoes, Z tetrominoes and Box tetrominoes. Prove that at least 4n − 1 L triominoes must be used.

Any ideas how can I solve this? Any help will be appreciated

Answer 1

Colour the cells of every other row of the board alternating blue and red. Cover the cells of the remaining rows alternating green and yellow (such that the first column is alternating blue and green). Then each tetromino covers one cell of each colour, while the triomino covers three different colours.

How many cells are there of each? There will be $n^2$ blue, $n(n-1)$ red and green, and $(n-1)^2$ yellow cells. To make up for this discrepancy, you need $2n-1$ triominoes that don't cover yellow, $n$ that don't cover red and $n$ that don't cover green.