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In how many ways can we choose two squares from a 8 by 8 chessboard given we cannot choose two squares that are in the same row or column?

There are 64c2 possible ways of selecting 2 squares on board. Subtract 8c2 *16 to subtract the options where there are two in the same row or column. Is this correct?

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  • $\begingroup$ What do you mean "so that no two..."??? You are choosing only two to begin with!!! $\endgroup$ Commented Oct 19, 2016 at 11:32
  • $\begingroup$ As in you may select any 2 squares, except those that are in the same column and row $\endgroup$
    – grigori
    Commented Oct 19, 2016 at 11:52
  • $\begingroup$ Well, this is confusing. You may as well replace "no two are" with "they are not". In any case, your answer looks correct to me. $\endgroup$ Commented Oct 19, 2016 at 11:56
  • $\begingroup$ I've edited it does it make sense now? $\endgroup$
    – grigori
    Commented Oct 19, 2016 at 12:00

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Almost. The number of possibilities with two in one row or column are a little bit wrong: Fixing one square there are only 15 elements in the corresponding row/column the other one can not take, not 16. Which one did you count twice?

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  • $\begingroup$ What OP did is $(\text{number of rows }+\text{ number of columns})\times(\text{ choose }2\text{ out of the }8\text{ squares in the chosen row or column})$. $\endgroup$ Commented Oct 19, 2016 at 11:44
  • $\begingroup$ Your suggestion (assuming that you meant $64\cdot15$) counts duplicates. $\endgroup$ Commented Oct 19, 2016 at 11:46
  • $\begingroup$ if you fix one square aren't there 14 others in the same row and column that it can't take? So -14*8? $\endgroup$
    – grigori
    Commented Oct 19, 2016 at 11:50
  • $\begingroup$ there aren't any that we counted twice though? As two squares cannot be in the same row and column $\endgroup$
    – grigori
    Commented Oct 19, 2016 at 12:01
  • $\begingroup$ No, there should be 15 as you also have to rule the case out that you pick the same square twice. This case is not ruled out when counting $64^2$ so you better do it at some place. $\endgroup$
    – Dirk
    Commented Oct 19, 2016 at 12:12

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