# A Combinatorics question

Gold Member
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

Why are you multiplying by 2?

disregardthat
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.

Gold Member
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

CRGreathouse
Homework Helper
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.

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.

disregardthat