Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
搜尋書籍內容
第 1 到 3 筆結果,共 88 筆
第 185 頁
... problem 83a ( with n ) , the number of arrangements of 2n customers ( or points ) which satisfy our hypothesis is ( n + 1 ) , we have m = 2n n 1 2n F = n + 1 n We have thus obtained a new solution to problem 54 . Another solution to problem ...
... problem 83a ( with n ) , the number of arrangements of 2n customers ( or points ) which satisfy our hypothesis is ( n + 1 ) , we have m = 2n n 1 2n F = n + 1 n We have thus obtained a new solution to problem 54 . Another solution to problem ...
第 191 頁
... problem . Remark . Problems 54 and 84b are special cases of the following more general problem . kn points on the circumference of a circle are given . In how many ways can these points be divided into n groups of k points each in such ...
... problem . Remark . Problems 54 and 84b are special cases of the following more general problem . kn points on the circumference of a circle are given . In how many ways can these points be divided into n groups of k points each in such ...
第 230 頁
... problem 83a , the answer to problem 54 is obtained immediately from this , Inasmuch as the answer to problem 53b can be derived from the answer to problem 54 ( compare with the solution to problem 54 ) , the reasoning indicated here ...
... problem 83a , the answer to problem 54 is obtained immediately from this , Inasmuch as the answer to problem 53b can be derived from the answer to problem 54 ( compare with the solution to problem 54 ) , the reasoning indicated here ...
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 points A1 polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose total number triangle unfavorable values vertex vertices