# 8 rooks on a chess board mustn't attack each other - not on white main diagonal

Let's assume we put 8 rooks on a chess board but we are not allowed to place them on the main diagonal with the white squares. In how many ways can I arrange them so that they cannot attack each other.

My approach is:

You can put the first rook in 8 different rows of the chess board and you can combine them with 7 columns. The second rook can be placed in 7 different rows and you can choose out of 6 columns and so on. This leads to $$8!7!$$ different possibilities. You then have to eliminate $$8!$$-many possibilities as the rooks are not distinguishable. So I get $$7!$$ possibilities.

I think this must be wrong although the reasoning seems legit to me?!

Where is my mistake?

• The question asks for the number of derangements of $\left\{1,2,\ldots,8\right\}$. You can easily find correct solutions to it. As for yours, I'm afraid I don't understand how it goes, so I cannot say where it goes wrong... – darij grinberg Nov 25 '19 at 23:37
• What happened when you tried your method on a $4\times4$ chessboard and compared your answer with the actual count? – bof Nov 25 '19 at 23:59

The flaw in the reasoning is because you write "the second rook can be placed in $$7$$ different rows and you can choose $$6$$ different columns and so on" which is not actually correct - and should stand out as a red flag because no justification is given to a pretty non-obvious and important fact in your argument. Also, your "and so on" hides what happens to the last rook in your argument, which you would say can be put into $$1$$ row and $$0$$ columns - which is clearly wrong!
Suppose we put the coordinates on the grid ranging from $$(1,1)$$ to $$(8,8)$$ where the diagonal in question is those points of the form $$(n,n)$$. If you put the first rook at $$(1,2)$$, your claim is that there are $$42$$ valid positions for the second rook - but this is not so! More specifically, you claim that we can fix the first coordinate in $$7$$ ways and then will have $$6$$ choices for the second coordinate - but this does not hold. In particular, if we choose the first coordinate for the second rook to be $$2$$, we find that all positions $$(2,x)$$ are legal except for $$(2,2)$$ - which is both attacked by the the first rook and on the main diagonal. Oops - so there are actually $$43$$ valid positions for the second rook!