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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 Solution? | Comment 1 of 5

Well, here's how I did it with a pencil and paper, there's probably a faster way:

Think of the board as a series of "rings" of squares.  The first ring is the squares on the edge of the board (A1, C1, F8, etc), then the next ring are the squares one away from the edge (B4, D7, E2, G6, etc)... 

If one queen is placed in any of the edge squares, there are 21 places you could place the other queen so that they are attacking each other.  If one queen is placed in one of the second ring squares, there are 23 places you could place the other queen so that they are attacking.  If one queen is places in one of the third ring squares, there are 25 places you could put the other queen, and finally if one queen is in one of the center four squares, there are 27 places to put the other queen so that they are attacking.

If you just look at a quarter of the board, there are 7 outer ring squares, 5 second ring, 3 third ring and 1 inner ring, so the number of possibilities are:

7*21 + 5*23 + 3*25 + 27 = 364.  Since that is a quarter of the board, we would have to multiply that by four to get 1456 ways two queens could be placed so that they are attacking each other.

Note that this is assuming the two queens are distinguishable - for example, a white queen on A1 and black queen on C1 is different than a black queen on A1 and white queen on C1.  If this is not the case, then the possibilities are halved, to 728.

In the spirit of the problem, I did all the math without a calculator, it's easy enough to do, although I'm not sure if there is an even easier way to figure this out that Jer is looking for.


  Posted by tomarken on 2006-05-19 10:04:29
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