Placing 8 identical rooks on the chessboard so that none of them attacks each other and none of them occupies the long diagonal

Question

In how many ways can we place $8$ identical rooks on a chessboard so that none of them attacks each other and none of them stands on the long diagonal A1 - H8?

I thought about the inclusion-exclusion principle, however, it is hard to determine how many possible movements each of these rooks will have - it depends on the choice of the one to rearrange.
Also, I made some observations on the chessboard and noticed that it is not possible to have an odd number of rooks out of the long diagonal whereas the rest of them occupy it - it may be of some help to the inclusion-exclusion but I can't link it together.

Any hints will be most appreciated.

Answer 1

Let $T$ denote the set of possibilities to place the rooks on the board in such a way that none of them attacks another.

For $L\in\{A,B,C,D,E,F,G,H\}$ let $L$ denote the subset of possibilities characterized by a rook on diagonal $A1-H8$ in row $L$.

Then to be found is:$$|A^{\complement}\cap\cdots\cap H^{\complement}|=|T|-|A\cup\cdots\cup H|=8!-|A\cup\cdots\cup H|$$

Then inclusion/exclusion and symmetry come in and the cardinality of an intersection of $k$ distinct elements from $\{A,B,C,D,E,F,G,H\}$ is $(8-k)!$.

This leads to:$$8!-|A\cup\cdots\cup H|=\sum_{k=0}^{8}\left(-1\right)^{k}\binom{8}{k}\left(8-k\right)!=14833$$possibilities.

Answer 2

Using Inclusion-Exclusion is one of the ways to approach this (and is good practice for inclusion-exclusion). Consider by the number of rooks that are on the diagonal.

Your total number of possibilities will be 8! (from here start excluding possibilities)

To see this consider that one rook will have to be placed in each row. For the first row you have 8 choices, for the second it can't go in the same column as the first so you have 7 choices and so on.

Try Inclusion-Exclusion from here and let me know if you need any more hints

Answer 3

Edit: this method fails to account for what happens when you choose the top row for picks after the first (see the first comment). It needs some adaptation.

A trick for approaching this:

• Place the first rook anywhere in the first column other than on the main diagonal. 7 choices.
• Go over on the row from your first choice to the main diagonal. Place the 2nd rook on that column. There are still 7 choices.
• Go over on the row from the 2nd rook to the main diagonal, and place the third rook in that column. Now there are 6 choices, as the first rook restricts a spot on this column.

Continue with this, always placing new rooks on the column where the previous rook attacks the main diagonal. Each new rook will have 1 few places available than the one before. Thus there will be $7\cdot 7! = 35280$ ways it can be done.