Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
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第 1 到 3 筆結果,共 24 筆
第 83 頁
... combining each of the two possible choices for the first rectangle with each of the two possible choices for the second , each of the two possible choices for the third , . . . , and each of the two possible choices for the ( n − 2 ) ...
... combining each of the two possible choices for the first rectangle with each of the two possible choices for the second , each of the two possible choices for the third , . . . , and each of the two possible choices for the ( n − 2 ) ...
第 145 頁
... Combining each of the 12 possibilities for the first person with each of the 12 possibilities for the second person , we obtain 12.12 = 122 equally likely possible outcomes for the first two people . Similarly we will have 123 equally ...
... Combining each of the 12 possibilities for the first person with each of the 12 possibilities for the second person , we obtain 12.12 = 122 equally likely possible outcomes for the first two people . Similarly we will have 123 equally ...
第 146 頁
... Combining each of the ways of choosing which three passengers will sit in the first car with each of the 26 ways of seating the other six passengers in the remaining two cars , we get 9.8.7.26 a total of • 26 - favorable outcomes . It ...
... Combining each of the ways of choosing which three passengers will sit in the first car with each of the 26 ways of seating the other six passengers in the remaining two cars , we get 9.8.7.26 a total of • 26 - favorable outcomes . It ...
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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 given Hence inclusion and exclusion intersection k-gons knights L₁ 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 T₂ total number triangle unfavorable values vertex vertices