# Last combinatorics question

In how many different ways can we place 8 identical rooks on a chess board so that no two of them attack each other?

• Lol at your title. – Q.matin Mar 24 '13 at 6:34

HINT: The chessboard has $8$ rows; number them $1$ through $8$. Let the rook in Row $1$ be in Column $c_1$, the rook in Row $2$ in Column $c_2$, and so on. The numbers $c_1,\dots,c_8$ must all be different and can be any arrangement of the $8$ column numbers.
Since rook in the $(i,j)$-th position can attack row $i$ and column $j$, for $i,j = 1,2,\dots,8$, it is equivalent to having the rooks on the diagonal. Since there are $8$ diagonal positions, the problem then becomes the number of ways of arranging the $8$ rooks, i.e. $8!$
You can only have one rook per column. Therefore for each column there is one rook. Call the rook in the first column $R_1$ in the second $R_2$ and so on. Therefore you must assign a row from $1$ to $8$ to each $R$ without repeating. There are $8!$ Ways to do so which is the same as $40320$.