Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 77 頁
... column or one rook in each row . For otherwise there would be a row and a column , neither of which contained any rooks ; and the square common to this row and column would not be controlled by any of the rooks . Conversely , if there ...
... column or one rook in each row . For otherwise there would be a row and a column , neither of which contained any rooks ; and the square common to this row and column would not be controlled by any of the rooks . Conversely , if there ...
第 94 頁
... column number . The remaining k queens lie in the ( k + 1 ) st through 2k - th columns ; the column number of the square in which one of these queens lies is thus of the form k + s , where s is a positive integer at most equal to k ; it ...
... column number . The remaining k queens lie in the ( k + 1 ) st through 2k - th columns ; the column number of the square in which one of these queens lies is thus of the form k + s , where s is a positive integer at most equal to k ; it ...
第 95 頁
A. M. I︠A︡glom, Isaak Moiseevich I︠A︡glom Basil Gordon. Fig . 42 - column lies in the 5th row , that in the 4th column lies in the 7th row , etc. ) . In the n / 2 - 3 columns starting with the ( n / 2 + 3 ) rd and ending with the ( n ...
A. M. I︠A︡glom, Isaak Moiseevich I︠A︡glom Basil Gordon. Fig . 42 - column lies in the 5th row , that in the 4th column lies in the 7th row , etc. ) . In the n / 2 - 3 columns starting with the ( n / 2 + 3 ) rd and ending with the ( n ...
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A₁ A₂ An+m arrangements b₁ B₂ binomial coefficients binomial theorem bishops black squares C₁ chessboard chord circle coefficient color column compute the number Consequently consider corresponding customers denote determine the number diagonals digits dihedral angle divided divisible draw equally likely possible equation equidistant equivalence classes exactly example experiment favorable outcomes follows formula G₁ given Hence inclusion and exclusion intersection k-gons knights length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks S₁ segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices