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There are $n^2$ squares disposed in a bigger n by n square, like a grid. Something like [this][1]
The question is: how many rectangles can be formed using those squares?
I saw this problem in a [blackpenredpen][2] (youtube maths channel) and decided to try and figure it out by myself, as I had an idea.
The idea I had was to calculate the number of possible diagonals that could be made from picking any two squares in the grid, since one diagonal is all it takes to define a rectangle. It made sense to divide by 2, since $n \choose 2$ would yield 2 diagonals for each rectangle.
So, I tried $\frac{n \choose 2}{2}$, but the results I got didn't match with the correct ones (I know the correct result, as it can be obtained by other methods, like the ones presented in the [video][2]).
I had really high hopes for this idea and was disppointed when I got an incorrect result.
I toyed and fiddled a little with the results I got, but to no avail. I did find out that for $n=8$ the difference between the result I got and the correct one was $7!$, which I found suspicious.
Can anyone help me figure out why my reasoning is not yielding the correct results?
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Your idea would actually yield $\frac12\binom{\color{red}{n^2}}2$, not $\frac12\binom{n}2$. But it doesn’t quite work, because picking the squares at the ends of a diagonal doesn’t count every rectangle twice: the rectangles that consist of squares in a single row or a single column are counted only once each by this method. – Brian M. Scott Oct 26 '20 at 17:31
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A few issues with your method: if you are choosing 2 squares from $n^2$ there should be $\binom {n^2}2$ choices; one block squares cannot be chosen since their diagonals only exist within one square; if your rectangle is one block thick, there is only one way to choose 2 squares generating its diagonal, instead of the usual 2. – player3236 Oct 26 '20 at 17:32