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So, there are alot of questions about tiling in this forum but I could not find this exact question.

I am trying to find out the number of possible "tile configurations" in an $n\times n$ grid where the tiles are $1\times 1$ and there are k less than or equal to $n^2$ of them. I have link down below with the "tile configurations" for a $2\times 2$ grid.

I don't know, but I feel like I am missing some simple way of approaching the problem. I've been trying to frame it in the context of the combination formula, but I feel like my intuition is lacking... Anyway, if someone could give me a hint that would be much appreciated.

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    $\begingroup$ Where does 17 come from? $\endgroup$ – user10354138 Jun 17 at 8:33
  • $\begingroup$ You missed one for $k = 2$ (all black ones on top), and you haven't counted $k = 0$ (the answer becomes much nicer if you do). $\endgroup$ – Arthur Jun 17 at 8:35
  • $\begingroup$ Hmm, for some reason the picture doesnt show ll configurations. There's supposed to be one more for k=1, k=2, and k=3, to left of the picture... $\endgroup$ – Victor Galeano Jun 17 at 8:52
  • $\begingroup$ Is this not "$k$ choices for the first tile, $k$ choices for the second tile, ..., $k$ choices for the $n \times n^\text{th}$ tile, so $k^{n^2}$ tilings"? If not, why not? $\endgroup$ – Eric Towers Jun 17 at 9:11
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For any integer $k \le n^2$ there are $\binom{n^2}{k}$ ways to pick $k$ tiles from the $n \times n$ grid. If you want to count all but $k \in \{0,n^2\}$, you will receive $$ \sum_{k=1}^{n^2-1}\binom{n^2}{k} = 2^{n^2} - \binom{n^2}{0} - \binom{n^2}{n^2} = 2^{n^2} - 2 $$ In your drawing, you missed "2 on top". Summing up the numbers for $k \in \{1,2,3\}$ you receive $$ 4 + 6 + 4 = 14 = 16 - 2 = 2^{2^2} - 2 $$

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