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$8$ rooks are randomly placed on different squares of a chessboard. A rook is said to attack all of the squares in its row and its column.

Compute the probability that every square is occupied or attacked by at least $1$ rook. You may leave unevaluated binomial coefficients in your answer.

I understand this problem but don't know how to start it. I know that you have to use PIE(Principle of Inclusion & Exclusion) in order to solve it, but other than that, I have no idea.

  • Duplicate? http://math.stackexchange.com/questions/1321371/rooks-attacking-every-square-on-a-chess-board – Robert Z Aug 05 '16 at 17:23
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    This post presents little if any indication that the problem was considered before posting. The phrasing of how "unevaluated binomial coefficients" can be left in the answer smacks of passing through an assignment undigested, and will likely receive a less sympathetic treatment than one where some effort or interest in the problem is shown. – hardmath Aug 06 '16 at 02:35

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Suppose that a row has no rook, then every column must have a rook, and vice-versa .

Therefore the number of good configurations is equal to the number of configurations that have a rook in every column, plus the number of configurations that have a rook in every row, minus the number of configurations that have a rook in every same row and every column.

How many configurations have a rook in every column? they are clearly $8^8$, for each column pick the position for the row.

How many configurations don't have a rook in every column and every row ? They are clearly $8!$, the you need to do the same as in the previous case, but without repeating.

Therefore there are $8^8+8^8-8!$ good configurations.

The total number of configurations is clearly $\binom{64}{8}$

The probability is hence $\frac{2\times 8^8-8!}{\binom{64}{8}}$

Asinomás
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