A Combinatorics question

  • Thread starter ShayanJ
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  • #1
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Find the number of ways of placing two knights in a chessboard that they can threaten each other.

I tried to solve it like this but it was wrong because the answer was not among the four options.
I wrote the number of squares that that the knight can threaten on every square that we place it.Then multiplied the number of squares with the same number by the number in it.then calculated the sum of the results and multiplied it by two.But as I said it was wrong although I'm still sure its right.

thanks
 

Answers and Replies

  • #2
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Why are you multiplying by 2?
 
  • #3
disregardthat
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Well, for each square, find the number of ways a knight can threated the square. After you have counted the sum of ways for all squares, you realize you have counted each way twice. Divide the total number by 2, and there you go.

Tell us your way of solving the problem.
 
  • #4
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I thought like below:
I should at first find the number of ways that I can place a knight on the chess board then multiply it by the number of squares it threatens when placed in a square then multiply it by two because I can do that with picking the white knight first or the black knight first.But as you know the number of squares a knight can threaten differs in differente areas of board.so I found the ways for each square seperatly.And the rest is obvious.And I don't think any thing is counted twice hear!Could you explain more?
thanks
 
  • #5
CRGreathouse
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And I don't think any thing is counted twice hear!Could you explain more?
thanks
Consider a chess board with only two squares such that a knight on one threatens the other. The count is two: both squares threaten one other square. But there's only one configuration where the knights attack each other: the configuration where both knights are on the only two squares.
 
  • #6
Consider a chess board with only two squares such that a knight on one threatens the other. The count is two: both squares threaten one other square. But there's only one configuration where the knights attack each other: the configuration where both knights are on the only two squares.
No, the knights have to be of opposite colours or they don't attack each other.
 
  • #7
disregardthat
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There was a slight confusion here. I assumed that color was irrelevant and any two knights were opponents. If you start out with a black knight, and for each square count the ways it can be threatened by a white knight you have counted all possibilities. There is no need in this case, as far as I can see, to multiply or divide by 2.
 
  • #9
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So you people mean my solution is all right except the part I multiply the number by 2?
 

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