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.
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 $$ 4 + 6 + 4 = 14 = 16 - 2 = 2^{2^2} - 2 $$