# Combinatorics. Rook Placing

Suppose that we want to place 8 non-attacking rooks on a chessboard. In how many ways can we do this if the 16 most ‘northwest’ squares must be empty? How about if only the 4 most ‘northwest’ squares must be empty?

1 Question : 4*3*2*1*4*3*2*1 = (4!)^2 Let's begin placing with 'northeast' corner => 4*3*2*1, then let's place remaining 'southwest' corner => 4*3*2*1

2 Question6*5*6*5*4*3*2*1 = 6*5*6! The same strategy. First place 'northeast' corner => 6*5, then 'southwest' corner => 6*5*4*3*2*1

I cannot find answers in the book or in internet, that is why are my solutions correct? If not, specify. Thank you.

• I think $(4!)^2$ and $(6\times 5)^24!$ are correct answers, but the second does not equal $8!$ (as you suggest). – drhab Nov 7 '18 at 16:36
• I edited it and it is (6*5)*(6!) – stackoverload Nov 7 '18 at 16:37
• I think that is correct, but why is it followed in your question by: $\cdots=8!$? – drhab Nov 7 '18 at 16:40
• Thank you. Now I will correct it. – stackoverload Nov 7 '18 at 16:43
• I suppose in part 2, "northeast corner" means the remaining part of the top two ranks (rows of squares) after removing 4 squares from the northwest, and "southwest corner" means the entire bottom 6 ranks. Otherwise this looks fine now. – David K Nov 7 '18 at 16:57

To place eight rooks in the case with the NW quadrant (rows 1-4, columns a-d) missing: - you must place four rooks in the NE quadrant (rows 1-4, columns e-h). There are 4! ways to do this. - you must then place four rooks in rows 5-8 - but you can't place them in any of columns e-h, so you essentially must place four rooks in the the 4-by-4 square of rows 5-8, columns a-d. There are 4! ways to do this. This gives $$4!^2$$ as the answer.
If instead you have a 2-by-2 NW corner missing (say a1, a2, b1, b2): - you must first place two rooks in rows 1 and 2. There are $$6 \times 5$$ ways to do this. - now you must place six rooks in rows 3-8, but there are two columns struck out (the columns in which you places the rooks in the first two rows) so there are $$6!$$ ways to do this. This gives $$6 \times 5 \times 6!$$ as the answer.
As a sanity check, if you have a 1-by-1 NW corner missing (say a1) this approach gives $$7 \times 7!$$. So the probability that a random arrangement of rooks doesn't contain a rook in a1 is $$(7 \times 7!)/8! = 7/8$$, and the probability that a random arrangement of rooks does contain a rook in a1 is therefore $$1 - 7/8 = 1/8$$. This is what you'd expect by symmetry.