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Given 8 rooks:

1) what is the probability that 3 rooks are placed on black squares and 5 on white squares?

2) what is the probability that there's only one rook on a each row and column?

I think that in total there can be $\binom{64}{8}$ possible positions for the rooks because there're 64 squares in a chessboard. Then:

1) there're $\binom{32}{3}$ way to place 3 rooks on black square and there're $\binom{32}{5}$ ways to place 5 rooks on white squares. Thus in total there're $\frac{\binom{32}{3}\binom{32}{5}}{\binom{64}{8}}$ ways.

2) this is an equivalent of permutation of 8 objects in a row. 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 $\frac{8!}{\binom{64}{8}}$.

Please let me know if my logic is correct.

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  • $\begingroup$ I think it's good! I can't find any gap. $\endgroup$ – user398842 Dec 15 '16 at 10:36
  • $\begingroup$ You are getting correct answers or not? $\endgroup$ – Kanwaljit Singh Dec 15 '16 at 14:06
  • $\begingroup$ For the sake of computing probability, you need to take identical arrangements into account. What do I mean? Well, you are correct in the assumption that there are $\binom{64}{8}$ different ways to arrange $8$ identical rooks on a chessboard with $64$ different squares. However, in order to compute the probability of some specific type of arrangements, you need to account for all arrangements, and consider any pair of identical arrangements as distinguishable. In this case, think of the rooks as different. $\endgroup$ – barak manos Dec 24 '16 at 21:46
  • $\begingroup$ So the answer to the first question, for example, is $\dfrac{\frac{32!}{3!}\cdot\frac{32!}{5!}}{\frac{64!}{8!}}$. $\endgroup$ – barak manos Dec 24 '16 at 21:46

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