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Mutually disagreeing queens. (Posted on 2006-05-19) Difficulty: 2 of 5
In how many ways can two queens be placed on a chessboard such that they are mutually attacking each other?

Show how to solve this problem quickly with only pencil and paper (not even a calculator).

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Solution I did it my way. | Comment 4 of 6 |

I solved this problem by recognizing a chessboard has symmetry and looked for the number of positions the queens could be in by counting each orthoganal and diagonal positions that could be attacked by one queen in each square of one quadrant. 

The seven positions on the outside edge of the quadrant each can attack 21 squares;  the five positions on the squares one square in each can attack 23 squares;  the three positions on the squares two squares in each can attack 25 squares; and, the single position on the innermost square can attack 27 squares.

A queen in one quadrant then can attack 7*21+5*23+3*25+1*27 squares, i.e.,  364 squares.  As there are four quadrants, 364*4 = 1456.

If the "identity" of the queen is not factored in, i.e., it does not matter which queen was in which location pair, 1456 would be divided by 2, giving 728.


  Posted by Dej Mar on 2006-05-19 23:18:13
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