$8$ 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 divided the question into cases, where Case $1$ is when there're only one rook in every row and column (which can be done in $8!$ ways).

But then, when I get to Case $2$ (which is when there're $2$ rooks in every row or column) I'm stuck. There're $\dbinom{8}{2}$ choices for choosing the squares in which the rooks will be, $8\cdot 7$ choices for those two rooks ... then 8C2 again for the next row, but then I overlap having more than one rook in a column ... and it's all confusing!


You should be using in inclusion-exclusion. Count the ways they fill all the rows, add the ways they fill all the columns, and subtract the ways they fill both because you counted them twice.

  • $\begingroup$ Yes. I agree with this. Though, to be pedantic, it would be a good idea to show that the only way to fulfil the condition in the question is if either all $8$ rows or all $8$ columns (or both) each contain a rook. $\endgroup$ – N. Shales May 8 '18 at 16:26

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