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 8 筆
第 99 頁
... knight . It follows from this that no more than four knights can be arranged in one of these rectangles in such a way that none of them lies on a square controlled by another . Therefore , the total number of knights which can be ...
... knight . It follows from this that no more than four knights can be arranged in one of these rectangles in such a way that none of them lies on a square controlled by another . Therefore , the total number of knights which can be ...
第 100 頁
... knights in that rectangle . therefore see that the bottom two squares of the first rectangle cannot simultaneously be occupied . If there were no knights in the bottom two squares of the first rectangle , then there would have to be knights ...
... knights in that rectangle . therefore see that the bottom two squares of the first rectangle cannot simultaneously be occupied . If there were no knights in the bottom two squares of the first rectangle , then there would have to be knights ...
第 101 頁
... knights in each of the second and third rectangles ( fig . 47f and g ) . But if these two knights lie in different columns ( that is , if they lie on squares of the same color ; see fig . 47h ) , then the two remaining knights can only ...
... knights in each of the second and third rectangles ( fig . 47f and g ) . But if these two knights lie in different columns ( that is , if they lie on squares of the same color ; see fig . 47h ) , then the two remaining knights can only ...
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常見字詞
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