Thread: Chess probelm (combinations)

Standard chess board.

1) How many ways to place 8 indistinguishable rooks on the chess board so that no rook is able to be attacked by another?
2) How many ways to place 8 indistinguishable rooks on the chess board so that each unoccupied square is able to be attacked by at least one rook?

I was thinking about it some more.

1) Is it just 8!? For the first row, there are 8 choices. For the second row, there are 7 choices, etc...
2) Is it just 8^8? Every row has to be hit, but now it doesn't matter if a row is repeated. First row has 8 choices, second row has 8 choices, etc...

I feel like these answers are too easy. Am I missing something?

Originally Posted by splash

Standard chess board.
1) How many ways to place 8 indistinguishable rooks on the chess board so that no rook is able to be attacked by another?
2) How many ways to place 8 indistinguishable rooks on the chess board so that each unoccupied square is able to be attacked by at least one rook?

A rook may attract only column-wise or row-wise.
So for part (A) you must place the rooks so that each column and each row has exactly one rook in it.

For (B), you must place one rook in each row OR place a rook in each column.

Originally Posted by Plato

A rook may attract only column-wise or row-wise.
So for part (A) you must place the rooks so that each column and each row has exactly one rook in it.

For (B), you must place one rook in each row OR place a rook in each column.

Yes, I understand that. This led me to my conclusions in the reply above. Are there any flaws in my logic?

Assuming you had the board started off with rooks at a1, b2, c3, d4, e5, f6, g7, and h8. You will be able to move the rook at a1 through all of the squares in its column. For example rook a1 goes to a2 while moving rook b2 to b1. Then rook at a2 to a3 while moving rook at b3 to b2. In this way the rook in the a column walks up through all its 8 squares. Repeat this for each of the 8 rooks. Because of the constraints, where each move of a rook along its column forces a specific and only one compensating move, I think the actual number of positions is 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 positions, that is, 8 ^ 2, or 64, which was one of your suspicions.