How countless distinct rectangles , v sides ~ above the network lines of the checkerboard and also containing at the very least 4 black color squares , deserve to be attracted on the checkerboard ?

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

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My concern is : Is over there a more clever method to settle this trouble , there is no summing or subtracting many cases? The total number of rectangles $n\times m$ in the board is $2(9-n)(9-m)$ if $m$ and also $n$ space different and also $(9-n)^2$ otherwise.

The rectangles that do not have 4 black squares space 1x1, 1x2, ..., 1x6, 2x2 and also 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 variety of rectangles in the board is given by $$\sum_k=1^8(9-k)^2+2\sum_j=1^7\sum_k=j+1^8(9-j)(9-k)=\left(\sum_k=1^8k\right)^2=1296$$ To conveniently count the variety of rectangles the any size: note that any kind of rectangle is uniquely identified by a left and a ideal horizontal net line, and a left and also right vertical gridline. There are 9 horizontal gridlines, therefore you need to select 2 different ones the end of those: $9 \choose 2$. Likewise, there room $9 \choose 2$ possible pairs of vertical gridlines. So, you have $9 \choose 2 \times 9 \choose 2$ feasible rectangles total: $36 \times 36 = 1296$ Thanks for contributing an answer to centregalilee.comematics stack Exchange!

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