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 5 筆
第 121 頁
... equivalence relation . A set in which an equivalence relation is defined can always be divided up into equivalence classes by putting C , and C2 in the same class whenever C1 ~ C2 . For a more complete description of this procedure ...
... equivalence relation . A set in which an equivalence relation is defined can always be divided up into equivalence classes by putting C , and C2 in the same class whenever C1 ~ C2 . For a more complete description of this procedure ...
第 122 頁
... equivalence classes . To do this it is essential to know how many colorings are equivalent to a given coloring C. First of all if C is a coloring in which all p sectors are painted the same color , then C is equivalent only to itself ...
... equivalence classes . To do this it is essential to know how many colorings are equivalent to a given coloring C. First of all if C is a coloring in which all p sectors are painted the same color , then C is equivalent only to itself ...
第 123 頁
... equivalence class of P. Our problem is to determine the number of equivalence classes . 0 Let Po be any polygon , and denote by Po , P1 , P2 , ... , Pp - 1 the polygons obtained by rotating P。 counterclockwise through angles of 0 ...
... equivalence class of P. Our problem is to determine the number of equivalence classes . 0 Let Po be any polygon , and denote by Po , P1 , P2 , ... , Pp - 1 the polygons obtained by rotating P。 counterclockwise through angles of 0 ...
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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 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