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This question is in my textbook:

What is the number of ways to put two non-identical knights on a chessboard so they attack each other?

My solution:

If two knights attack each other, they can be fit inside a $2 *3$ rectangle. There are $84$ ways to pick a $2 *3$ rectangle from the chessboard(horizontal and vertical) and there's $4$ ways to put $2$ non-identical knights inside such a rectangle so the answer is $4*84=336$.

But my textbook says it's $672$. I checked out this question and it said the number of ways to put $2$ attacking identical knights in a $n*n$ board is $4(n-1)(n-2)$ and substituting $8$ yields $4*7*6 = 168$. Note that since the knights are non-identical, we need to multiply $168$ by $2$ which results in the same answer as mine. So I'm fairly certain my answer is right and the textbook's is wrong but I wanted to make $100\%$ sure.

Thank you in advance!

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    $\begingroup$ I agree with your answer. $\endgroup$ Nov 12, 2019 at 20:49
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    $\begingroup$ On a $n\times n$ chessboard, a black and a white mutually-attacking knights can only be in opposite corners in a $2\times 3$ or $3\times 2$ rectangle. There are $(n-2)(n-1)$ ways to select a $3\times 2$ rectangle, so $2(n-2)(n-1)$ ways to select a $2\times 3$ or $3\times 2$ rectangle. Once such selection is performed, we still have to decide if our knights occupy the NE-SW diagonal or the NW-SE diagonal, then if they occur in the BW or WB order. Thus a total of $8(n-2)(n-1)$ ways, which equals $336$ if $n=8$, I agree with you. $\endgroup$ Nov 12, 2019 at 20:56
  • $\begingroup$ I agree with your analysis and the analysis of the other two comments. I offer a more pedestrian way of arriving at the same answer of 336. The # of 2 row x 3 column rectangles can be identified by the possible choices for the lower left hand corner of such a rectangle. There is a choice of 6 columns and 7 rows = 42 different squares that the lower left hand corner may occupy. This means that there are 42 possible 2 row x 3 column rectangles. By symmetry there must also be exactly 42 possible 3 row x 2 column rectangles. ...see next comment. $\endgroup$ Aug 18, 2020 at 15:55
  • $\begingroup$ Therefore, there are 84 possible (2 x 3 or 3 x 2) rectangles. Since the knights are construed to be different, pretend that one is white and one is black. In a (2 x 3 or 3 x 2) rectangle, the only way that two knights can attack each other is if they occupy "opposite corners". Select any one of the 4 corners in the rectangle to place the white knight. This fixes the position of the black knight. Thus, for each (2 x 3 or 3 x 2) rectangle, there are 4 possible configurations where the black and white knights attack each other. $\endgroup$ Aug 18, 2020 at 15:59

1 Answer 1

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Number of $ 2 * 3 $ rectangles is: $ (n - 2)(n - 1) + (n - 1)(n - 2) = 2(n - 1)(n - 2) $. In the $ 2 * 3 $ rectangle like this:

c1 c2 c3
c4 c5 c6

c1 horse atacks only c6, and c3 horse atacks only c4. But, since the horses are non-identical, we have $ 2 * 2 = 4 $ possible combinations.
Therefore, the answer is $ 8(n - 1)(n - 2) $.

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