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How many ways are there to place $n$ figures on an $n \times n$ chess board, if there are $n-k$ indistinguishable black figures and $k$ indistinguishable white figures in such a way that there is exactly one figure in each row?

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  • $\begingroup$ One figure in each row, but not necessarily each column, correct? $\endgroup$
    – MT_
    Commented Oct 12, 2015 at 16:25
  • $\begingroup$ Is the chessboard marked (i.e. admits algebraic notation) or not? This matters since in the first case the orientation of the board is unique, while in the later it is symmetric relative to inversions with respect to the main diagonals... $\endgroup$
    – Newbie
    Commented Oct 12, 2015 at 16:26

2 Answers 2

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Shouldn't that be $n^n\cdot{n\choose k}$ for obvious reasoins?

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  • $\begingroup$ The only thing to maybe consider and is unclear from the problem is if we may rotate the chess board $\endgroup$
    – MT_
    Commented Oct 12, 2015 at 16:28
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First try to choose the positions that would be occupied by the figures. From each row you can select only one position in $n$ ways. So the $n$ positions can be chosen in $n^n$ ways. Now distribute the figures in your chosen $n$ positions. Choose any $k$ using ${n \choose k}$ ways to place the white figures and use the rest for the white ones. Your answer would be $n^n{n \choose k}$

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