Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 1 到 3 筆結果,共 5 筆
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 . Let Po be any polygon , and denote by Po , P1 , P2 , ... , Pp - 1 the polygons obtained by rotating P , counterclockwise through angles of 0 ° , 360 ...
... equivalence class of P. Our problem is to determine the number of equivalence classes . Let Po be any polygon , and denote by Po , P1 , P2 , ... , Pp - 1 the polygons obtained by rotating P , counterclockwise through angles of 0 ° , 360 ...
第 192 頁
... equivalence relation , and therefore S is partitioned into equivalence classes as explained on p . 121 . Thus in the case m = 3 , n = 2 , the five sequences given above form one equivalence class . We will show that when m and n are ...
... equivalence relation , and therefore S is partitioned into equivalence classes as explained on p . 121 . Thus in the case m = 3 , n = 2 , the five sequences given above form one equivalence class . We will show that when m and n are ...
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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 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