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?
Question 1 : $4\cdot3\cdot2\cdot1\cdot4\cdot3\cdot2\cdot1 = (4!)^2$
Let's begin placing with 'northeast' corner $\Rightarrow$ $4\cdot3\cdot2\cdot1$, then let's place remaining 'southwest' corner $\Rightarrow$ $4\cdot3\cdot2\cdot1$
Question2: $6\cdot5\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1 = 6\cdot5\cdot6!$
The same strategy. First place 'northeast' corner $\Rightarrow$ $6\cdot5$, then 'southwest' corner $\Rightarrow$ $6!$
I cannot find answers in the book or in internet, that is why are my solutions correct? If not, specify. Thank you.
 A: Let's label the rows 1 through 8 and the columns a through h, as is standard in chess notation. 
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.
