Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 1 到 3 筆結果,共 13 筆
第 11 頁
... bishop ? Solve the same problem for an n × n chessboard . 36. Prove that for even n the following numbers are perfect squares : a . the number of different arrangements of bishops on an n × n chessboard such that no bishop controls a ...
... bishop ? Solve the same problem for an n × n chessboard . 36. Prove that for even n the following numbers are perfect squares : a . the number of different arrangements of bishops on an n × n chessboard such that no bishop controls a ...
第 78 頁
... bishops in such an arrangement cannot exceed 14 . On the other hand , 14 bishops can be arranged in the required way , as is shown , for example , in fig . 29. Hence this is the greatest number of bishops which can be arranged on an 8 x ...
... bishops in such an arrangement cannot exceed 14 . On the other hand , 14 bishops can be arranged in the required way , as is shown , for example , in fig . 29. Hence this is the greatest number of bishops which can be arranged on an 8 x ...
第 81 頁
... bishop lies on a square con- trolled by another , and the corresponding problem for the black bishops . But in the case of even n , the union of all the black squares on the board and the union of all the white squares are congruent ...
... bishop lies on a square con- trolled by another , and the corresponding problem for the black bishops . But in the case of even n , the union of all the black squares on the board and the union of all the white squares are congruent ...
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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