# In how many ways can four squares, not all in the same row or column, be selected from an $8 \times 8$ chessboard to form a rectangle?

I know this involves creating a combinatorial argument but this problem is really complex.

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For rectangles with sides parallel to the sides of the board, you just have to select two rows out of eight and two columns out of eight. Are you counting oblique rectangles, too?

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No oblique rectangles will not count –  user61646 Feb 10 '13 at 18:28
Then you have $8\cdot 7/2$ ways to choose the columns, and the same for the rows, making $28^2=824$ –  Ross Millikan Feb 10 '13 at 22:52
I'm guessing you meant 28^2 = 784 –  user61646 Feb 13 '13 at 3:17
@user61646: yes. –  Ross Millikan Feb 13 '13 at 3:39