# Combinatorics Problem involving chessboard and squares

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?

• What do you mean "so that no two..."??? You are choosing only two to begin with!!! – barak manos Oct 19 '16 at 11:32
• As in you may select any 2 squares, except those that are in the same column and row – grigori Oct 19 '16 at 11:52
• 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. – barak manos Oct 19 '16 at 11:56
• I've edited it does it make sense now? – grigori Oct 19 '16 at 12:00

## 1 Answer

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?

• What OP did is $(\text{number of rows }+\text{ number of columns})\times(\text{ choose }2\text{ out of the }8\text{ squares in the chosen row or column})$. – barak manos Oct 19 '16 at 11:44
• Your suggestion (assuming that you meant $64\cdot15$) counts duplicates. – barak manos Oct 19 '16 at 11:46
• if you fix one square aren't there 14 others in the same row and column that it can't take? So -14*8? – grigori Oct 19 '16 at 11:50
• there aren't any that we counted twice though? As two squares cannot be in the same row and column – grigori Oct 19 '16 at 12:01
• 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. – Dirk Oct 19 '16 at 12:12