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 43 筆
第 46 頁
... four pairs of vertical trihedral angles , the locus of all points equidistant from the given planes consists of four straight lines through P. 2 , 3 1 , 3 , 4 19 4 Now let П1 , П2 , П ̧ , П be the planes of the faces of the tetrahedron ...
... four pairs of vertical trihedral angles , the locus of all points equidistant from the given planes consists of four straight lines through P. 2 , 3 1 , 3 , 4 19 4 Now let П1 , П2 , П ̧ , П be the planes of the faces of the tetrahedron ...
第 99 頁
... four knights ( since we have 32 knights to dispose of and by the argument of part a , no more than four can be in any one section ) . Consider now how four knights can be arranged on the lower left - hand rectangle ( we will call this ...
... four knights ( since we have 32 knights to dispose of and by the argument of part a , no more than four can be in any one section ) . Consider now how four knights can be arranged on the lower left - hand rectangle ( we will call this ...
第 148 頁
... four games is 24 = 16 , since each game has two possible outcomes . These are all equally likely , since A and B are equally strong . The favorable outcomes are those in which B wins only one game out of the four ; this can happen in four ...
... four games is 24 = 16 , since each game has two possible outcomes . These are all equally likely , since A and B are equally strong . The favorable outcomes are those in which B wins only one game out of the four ; this can happen in four ...
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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