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You have 8 rooks. What is the probability of placing all 8 rooks on an 8 by 8 chess board with out one being able to hit each other? But there's a catch of course..one of the spaces is unavailable to be used. That being said, there could potentially be 2 rooks in a certain row or column. The unavailable space is 7 columns over and 7 rows down.

Without the catch it would be $\frac{8!}{64\choose 8}$

I came up with.. $\frac{8\cdot{7\choose 2}6\cdot6\cdot5\cdot4\cdot3\cdot2}{63 \choose 8}$ but I don't think I'm correct. Any advice/ideas?

tim
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  • Look at the blocked row and blocked column. Either they both have two rooks (count the ways) or one has two rooks (count the ways) or neither has two rooks (7765432*1 ways) – Empy2 Apr 15 '15 at 14:12
  • It matters where the blocked square is, I think. In the center? – mjqxxxx Apr 15 '15 at 14:18
  • The blocked square is 7 columns over and 7 rows down – tim Apr 15 '15 at 14:34
  • Your answer can't be right, because the numerator is three times as large as the numerator in the case where all squares are allowed. Deleting a square can only remove possibilities. – Ross Millikan Apr 15 '15 at 14:48
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    "Blocked" means that you can't place rooks there or that you can't place rooks there and two rooks that are on the same row of the blocked square can't hit each other? – Ant Apr 15 '15 at 14:55

2 Answers2

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There are ${64\choose8}$ ways to place the rooks when there are no restrictions. Exactly one eighth of these have a rook on the forbidden square. It follows that there are $${7\over8}\>{64\choose8}=3\,872\,894\,697$$ admissible placements of the rooks.

There are $8!$ good placements of the rooks when there are no restrictions. Exactly one eigth of these have a rook on the forbidden square, so that $${7\over8}\>8!=35\,280$$ good placements remain having no two rooks in one row or column.

We can have two rooks in row$_7$, but no two rooks in the same column. The first rook in row$_7$ can be placed in $6$ ways. Then we can choose the row remaining empty in $7$ ways, and finally we can place the remaining $6$ rooks such that a good configuration results in $6!$ ways. This makes for $$6\cdot 7\cdot 6!=30\,240$$ good configurations of this type, and the same number results when we start with col$_7$ having two rooks.

When we have two rooks in row$_7$ as well as in col$_7$ then we can place the first rook in row$_7$ and the upper rook in col$_7$ in $6$ ways each; then we can choose the row and the column remaining empty in $5$ ways each, and finally we can place the remaining $4$ rooks such that a good configuration results in $4!$ ways. This makes for $$36\cdot 25\cdot 4!=21\,600$$ good configurations of this type,

All in all there are $117\,360$ admissible good placements, so that the required probability $P$ computes to $$P={117\,360\over3\,872\,894\,697}\doteq3.03029\cdot10^{-5}\ .$$

When there are no restrictions the corresponding probability $P_0$ is given by $$P_0={8!\over{64\choose8}}\doteq9.10947\cdot10^{-6}\ ,$$ which is considerably smaller.

Christian Blatter
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The easiest way I see to count successes is to take the number of successes for the full chess board, $8!$ and note that $\frac 18$ of them use the forbidden square. That gives $7\cdot 7!$ successes. Another way to see this is to place a rook in the column with the forbidden square, which gives $7$ choices. Then the next rook also has $7$ choices and so on. The chance of success is $\frac {7\cdot 7!}{63 \choose 8}=\frac {560}{61474519}\approx 9.11\cdot 10^{-6}$ The probability is identical to the case with a complete board.

Ross Millikan
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