Placing 8 rooks onto a 8x8 chess board so no two share same row or column

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SUMMARY

The problem of placing eight identical rooks on an 8x8 chessboard without sharing rows or columns can be solved using combinatorial mathematics. The total arrangements for distinguishable rooks is calculated as 8! (factorial of 8), resulting in 40,320 unique configurations. For identical rooks, the arrangements reduce significantly, requiring division by 8! to account for indistinguishable placements, leading to a total of 1 configuration. The discussion also explores variations, such as placing rooks only on black squares, which introduces additional constraints.

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  • Familiarity with chessboard configurations and rook movement rules.
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nowimpsbball
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Homework Statement


In how many ways can eight identical rooks be placed on an ordinary 8x8 chessboard so that no two are in the same row or column? In how many ways, if each rook has a different color?



Homework Equations


I looked at the equation n choose k (nck), but I don't know if that would work


The Attempt at a Solution


I know each time you place a rook that the number of spaces to put the next rook goes down to the next perfect square...
(Choice, Available Places to put rook)
(1,64)
(2,49)
(3,36)
(4,25)
(5,16)
(6,9)
(7,4)
(8,1)

I thought it might be 64*49*36*...*1, but that seems like that would be too large a number over 1.625 billion...my intuition says that is way too big.

Any hints, tips, suggestions, answers for this problem? Am I thinking about it the right way? Thanks
 
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I believe that is correct if all of the rooks have different colors (and there are 8 colors). So they are distinguishable. What if they aren't? How many ways to assign 8 colors to 8 rooks in a given position?
 
From the question, since the Rooks are identical, I think we can assume that, in a given arrangement, even if we swap two Rooks, it wouldn't make a "new" arrangement.

Your answer "1625702400" is right if all the Rooks are distinguishable. So I'll give you a small exercise, and this would be for the first part, i.e. no Rooks are distinguishable (i.e. identical).

Consider a 4x4 square, where you have to arrange 2 Rooks, and then use the same logic you used, i.e. the first rook can be assigned any four squares, but the second only 1. Can you now draw these combinations? What does that tell you? Does things change if rooks are identical, i.e. "swapping" of rook doesn't matter?

So you have to do something with 1625702400, to give you the first part of your question. 1625702400 is a good answer for the second, I believe.
 
Is there an echo in here? Why are you reiterating the same thing I just said?
 
So i should divide the first answer by 8, cool, thanks.
 
nowimpsbball said:
So i should divide the first answer by 8, cool, thanks.

NO! Divide by 8!. Why?
 
Dick said:
NO! Divide by 8!. Why?

oh duh, because there are 8 different possibilities for each spot so 8!
 
If it were 8 possibilities for each of 8 spots, that would be 8^8. You aren't giving me a lot of confidence here...
 
what if the rooks were to be placed on only black squares?
would that give 576*(8)! ?
 

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