1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What do you mean "so that no two..."??? You are choosing only two to begin with!!! $\endgroup$ – barak manos Oct 19 '16 at 11:32
  • $\begingroup$ As in you may select any 2 squares, except those that are in the same column and row $\endgroup$ – grigori Oct 19 '16 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. $\endgroup$ – barak manos Oct 19 '16 at 11:56
  • $\begingroup$ I've edited it does it make sense now? $\endgroup$ – grigori Oct 19 '16 at 12:00
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What OP did is $(\text{number of rows }+\text{ number of columns})\times(\text{ choose }2\text{ out of the }8\text{ squares in the chosen row or column})$. $\endgroup$ – barak manos Oct 19 '16 at 11:44
  • $\begingroup$ Your suggestion (assuming that you meant $64\cdot15$) counts duplicates. $\endgroup$ – barak manos Oct 19 '16 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 Oct 19 '16 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 Oct 19 '16 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 Oct 19 '16 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.