Number of rectangles on a checkerboard with at least 4 black squares

Question

How many distinct rectangles , with sides on the grid lines of the checkerboard and containing at least 4 black squares , can be drawn on the checkerboard ?

I tried to solve the problem by counting all the possible rectangles on a 8x8 board and then taking away all the rectangles with less than 4 squares but this type of counting took me a lot of cases to deal with (1x1 , 1x2 , 1x3 , 2x3 , ecc...).

My question is : Is there a more clever way to solve this problem , without summing or subtracting many cases?

Answer 1

The total number of rectangles $n\times m$ in the board is $2(9-n)(9-m)$ if $m$ and $n$ are different and $(9-n)^2$ otherwise.

The rectangles that do not have four black squares are 1x1, 1x2, ..., 1x6, 2x2 and 2x3, and the half of 1x7.

This makes $$8^2+2\cdot 7\cdot 8+2\cdot 6\cdot 8+\cdots 2\cdot 3\cdot 8+2\cdot 8+7^2+2\cdot 6\cdot 5$$ $$=64+2\cdot8\sum_{k=3}^7k+16+49+60$$ $$=64+25\cdot16+16+49+60=589$$

The total number of rectangles in the board is given by $$\sum_{k=1}^8(9-k)^2+2\sum_{j=1}^7\sum_{k=j+1}^8(9-j)(9-k)=\left(\sum_{k=1}^8k\right)^2=1296$$

Answer 2

To quickly count the number of rectangles of any size: Note that any rectangle is uniquely defined by a left and a right horizontal grid line, and a left and right vertical gridline. There are 9 horizontal gridlines, so you need to choose 2 different ones out of those: $9 \choose 2$. Likewise, there are $9 \choose 2$ possible pairs of vertical gridlines. So, you have ${9 \choose 2} \times {9 \choose 2}$ possible rectangles total: $36 \times 36 = 1296$