Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
搜尋書籍內容
第 1 到 3 筆結果,共 32 筆
第 頁
... Answers and hints are given at the end of the book . For most of the problems the reader is advised to find a solution by himself . After solving the problem , he should check his answer against the one given in the book . If the answers ...
... Answers and hints are given at the end of the book . For most of the problems the reader is advised to find a solution by himself . After solving the problem , he should check his answer against the one given in the book . If the answers ...
第 229 頁
... answer to the problem follows easily from this . n + m m - ; the Second solution . Given the answer to the problem , it is not hard to verify its validity by mathematical induction . Third solution . Consider the n + m arrangements ...
... answer to the problem follows easily from this . n + m m - ; the Second solution . Given the answer to the problem , it is not hard to verify its validity by mathematical induction . Third solution . Consider the n + m arrangements ...
第 230 頁
... answer to problem 53b can be derived from the answer to problem 54 ( compare with the solution to problem 54 ) , the reasoning indicated here also solves problem 53b . b . The required number G , of ways is ( 2n + 2 ) ( 2n + 3 ) ...
... answer to problem 53b can be derived from the answer to problem 54 ( compare with the solution to problem 54 ) , the reasoning indicated here also solves problem 53b . b . The required number G , of ways is ( 2n + 2 ) ( 2n + 3 ) ...
其他版本 - 查看全部
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