1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think $(4!)^2$ and $(6\times 5)^24!$ are correct answers, but the second does not equal $8!$ (as you suggest). $\endgroup$ – drhab Nov 7 '18 at 16:36
  • $\begingroup$ I edited it and it is (6*5)*(6!) $\endgroup$ – stackoverload Nov 7 '18 at 16:37
  • 1
    $\begingroup$ I think that is correct, but why is it followed in your question by: $\cdots=8!$? $\endgroup$ – drhab Nov 7 '18 at 16:40
  • $\begingroup$ Thank you. Now I will correct it. $\endgroup$ – stackoverload Nov 7 '18 at 16:43
  • $\begingroup$ 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. $\endgroup$ – David K Nov 7 '18 at 16:57
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.