What is the maximum number of T-shaped polyominos (shown below) that can be put into a 6x6 grid without any overlaps? The blocks can be rotated.

Question

I just drew the figure and manually tried the question but I am wondering is there a way to do this problem via permutations and combinations.

PS: I got answer as 7.

Answer 1

(Fill in the gaps as needed. IF you're stuck, explain where you are stuck and show your work.)

Hint: The answer is

8

1. Find a configuration that has that many T's.

2. To prove that we can't do 9 T's, color the grid like a chessboard. Notice that there are 18 Black and 18 White squares. Notice that every T either covers 3 Black 1 White or 3 White 1 Black squares. Find a contradiction.

There is no non-negative integer solution to $a+b = 9, 3a+b = 18$, hence we can't place 9 tiles.

Answer 2

I filled the 6x6 square manually withou any calculation or considering possibility of other cases.

Answer 3

As noted above it's possible to tile the 6x6 square with 8 T-tetrominoes, with 4 squares left over. FWIW, there's a simple tiling of the 4 x 4 square, so there are trivial tilings of any 4n x 4m rectangle.

Here's a symmetrical 6×6 solution I found by hand:

(JPEG version)

I hacked my code from my answer to Tiling of a $9\times 7$ rectangle to solve this new problem. Due to the nature of the hack, I'm not totally confident that I've found all the solutions, but I managed to find 184, which include rotations and reflections. So the true answer is something higher than 184 / 8 = 23.

Anyway, here's a slow-motion GIF anim of those results. I suppose it would look nicer if the 4 empty squares were all the same colour, and a different colour to the tetrominoes, but I didn't feel up to hacking that part of the program too.