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Expert: - 6/1/2010


Question
How many ways are there to place two identical qeens on an 8 x 8 chessboard so that the queens are not in a common row, column, or diagonal?


I have started the problem, but I'm not sure if I am on the right track or not. Basically, I drew an 8 x 8 board and tested different positions.

For the four corners, the number of total ways is ((4)(42))
For the spaces that are adjacent to the corners, it is ((8)(42)). For the spaces 2 above the bottom corners and 2 below the top corners, it is ((8)(42)). For the spaces one diagonal above the corners, it is ((4)(40)). For the spaces two diagonal from the corner, it is ((4)(38)). For the spaces three diagonal from the corners, it is ((4)(36)). For the spaces two above (or below), one over the corners, it is ((8)(40)). For the spaces three above (or below), one across from the corners, it is ((8)(40)). Lastly, for the spaces three above or below the corner, two spaces over, it is ((8)(38)).

All of these numbers multiplied is 5.1692 x 10^20.
All of these numbers added (for example, (8 times 38) plus (8 times 40)) is 2240 ways.

I think that adding these numbers leads to the correct solution.

Is 2240 the right answer?

Thanks for your help!!
-Leanna

Answer

Counting
The following has lots of words, for that is what it took to explain.
I've heard a picture is worth a thousand words, so I attached a picture as well,
even though there are only around 550 words, for there are over 2,000 items
if the numbers are counted as well..

That is a good approach.  Let's check it out.  I've got a totally different way of approaching the problem that seems much clearer (to me anyway).  Picutre the first queen being in one of found locations: on the edge, one square from the nearest edge, two squares from the nearest edge, or 3 squares from the nearest edge.  I'll discuss the number in each near the end.

By placing the queen anywhere, there are always 7 squares in the same row and seven squares in the same column, and the one the queen is on for a total of 7+7+1=15 squares that the other queen can't go on.

Now if the queen is on the outer edge, there are 7 diagonals it can cover.  Adding on the 15 straigt squares to the 7 diagonals gives 22 ways given one queen is on the edge.
Outside: 64 - 22 = 42 squares for other queen for each choice for each choice for the first queen.

If the queen is one square in, there are 2 squares towards the outside with 7 away fromt the outsides, for a total of 9.  Since there are 15 squares for horinontal and vertical, that makes for a total of 9 + 15 = 24.
One square in: 64 - 24 = 40 squares for other queen for each choice for each choice for the first queen.

If the queen is two squares from any edge, there are 2 squares diagonally in each direction towards the edge that she can go to (for a total of 4, since there are two directions) and 7 others, which makes for 11 squares.  This makes for a total square count of 15 + 11 = 26.
Two squares in: 64 - 26 = 38 squares for other queen for each choice for each choice for the first queen.

If the queen is three squares from any edge, that only involves the 4 squares that are in the middle of the board.  The number of diagonals that the queen could get to would be 7 going one direction and 6 going the other of reach square.  This makes for a total of 13 diagonals. Adding in the 15 vertical and horizontal squares gives 28.
Three squares in: 64 - 28 = 36 squares for other queen for each choice for the first queen.

Now that we have the number of choices for how many squares the queen is from the edge,
for the last one, there are 4 squares in the center that are at least 3 from any edge.
For being two squares away, that is a 4x4 square, which is 4², but we have to take away the middle.  That is 4² - 2² = 16 - 4 = 12 squares.  For being one square away, that is 6² - 4² = 36 - 16 = 20 squares.  For being on the edge, that is 8² - 6² = 64 - 36 = 28 squares.

Taking note of all of these gives 4, 12, 20, and 28 squares with the options being
42, 0, 38, and 36, respectively.

That is, 4*42 + 12*40 + 20*38 + 28*36 = 168 + 480 + 760 + 1008 = 2416.

Now at the end, I notice that 2416 is close to 2240, yet reviewing what I've given you,
I'm not sure where the error would be.

Scotto

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