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 22 筆
第 40 頁
... plane ABC . Thus there is one and only one plane II equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B ...
... plane ABC . Thus there is one and only one plane II equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B ...
第 40 頁
... plane ABC . Thus there is one and only one plane ПI equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B ...
... plane ABC . Thus there is one and only one plane ПI equidistant from A , B , C , D with A , B , C on one side of it and D on the other side . By the same reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B ...
第 104 頁
... plane into two parts , the total number of parts after drawing the n - th circle is 2 + 2 + 46 +8 + ··· + 2 ( n − 1 ) • = 2 + 2 ( 1 + 2 + 3 + ··· + ( n − 1 ) ) = 2 + 2 n ( n 2 ― 1 ) n2 = n - n + 2 . 45a . n planes will divide 3 ...
... plane into two parts , the total number of parts after drawing the n - th circle is 2 + 2 + 46 +8 + ··· + 2 ( n − 1 ) • = 2 + 2 ( 1 + 2 + 3 + ··· + ( n − 1 ) ) = 2 + 2 n ( n 2 ― 1 ) n2 = n - n + 2 . 45a . n planes will divide 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