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$8$ rooks are placed randomly on an $8\times 8$ chess board.

What is the probability of having exactly one rook each row and each column?

I guess there is no meaningful order here?

Michael Hardy
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Ilona Shishov
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    You can put a rook in any of the $8$ positions in the first row.

    Then one in any of the seven remaining positions in the second row.

    Then one in any of the six remaining positions in the third row.

    $\ldots\text{ and so on }\ldots\qquad$

     – Michael Hardy Dec 16 '16 at 15:41

2 Answers2

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Hint. Eight (indistinguishable) rooks can be placed on an $8\times 8$ chess board in $\binom{64}{8}$ ways (we have to select $8$ positions among $8^2=64$).

Having exactly one rook each row and each column can be done in $8!$ ways (for each of the $8$ columns we choose e different row).

Robert Z
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Once we place a rook on row $x$ and column $y$, there remain $x-1$ and $y-1$ possibilities for future placements. Thus the answer is $\frac{8!}{\binom{64}{8}}$.

Kanwaljit Singh
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  • Your welcome. I think you have to accept answer you understand. So that in future no duplicate question again. And the person has same problem get help easily. – Kanwaljit Singh Dec 17 '16 at 16:42