# 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?

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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.

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Hint: Think about how many ways you can place the first rook on the board, then the second one, the third one, and so on. Then think about how many ways you can place them (in total) if the rooks were distinguishable.

That should get you going.

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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. $8!$

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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 $40320$.

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