0

I'm trying to calculate the probability that if 8 rooks are placed on a chessboard, they are placed in a fashion such that none of the rooks can capture another rook. So, no row or column can contain more than one rook.

I figure that the total number of positions they can be placed is C(64, 8). As for the number of places they can be placed without capturing one another, I modeled it as a set of vectors where each element is 1(where a rook is present), or a 0(no rook). So I need to find the number of sets of vectors such that the sum of all vectors is (1,1,1,1,1,1,1,1), that is, no value 1 shares a row or column.

The first vector is C(8,1), the second is C(7, 1) and so on. So I got:

8! / C(64, 8).

I don't feel this is correct though, so some help would be much appreciated.

Jaigus
  • 505
  • This is in fact correct. – JMoravitz Feb 20 '19 at 00:28
  • I get the same result. – Doug M Feb 20 '19 at 00:30
  • Thanks for the answers and comments. Also, to whoever voted to merge, the two questions are indeed very close however this is asking for the probability the pieces are arranged this way after random placement, while the other is asking for the number of possible arrangements. – Jaigus Feb 20 '19 at 01:10
  • 1
    The only additional step to take is dividing by the number of arrangements which is clearly $\binom{64}{8}$. The only difficulty for the problem should have been totally addressed by the linked duplicate. – JMoravitz Feb 20 '19 at 01:17

1 Answers1

1

You are correct. Another way of looking at this (which may be equivalent to yours) is as follows:

Consider a diagonal unit matrix of the format $8\times8$. Let the ones represent all the positions with rooks in them. All the possible configurations where no rooks can attack each other are obtained by permutating the matrix' rows/columns. Note that it is enough to count the number of permutations of rows only, because permutating the columns can give us the same configurations.

Finally, the number of configurations is equal to the number of such matrices, which is $8!$.

Haris Gušić
  • 4,217