# 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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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