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 34 筆
第 69 頁
... triangle of perimeter n . Thus the problem is similar to the preceding one , but differs from it in two respects . First , the required inequalities contain a < sign where there was a sign before . Secondly , solutions differing only in ...
... triangle of perimeter n . Thus the problem is similar to the preceding one , but differs from it in two respects . First , the required inequalities contain a < sign where there was a sign before . Secondly , solutions differing only in ...
第 114 頁
... triangle are parts of diagonals А , and AA of the n - gon ; our triangle is one of the four triangles into which the quadrilateral А12 , is divided by its diagonals . Thus each A4 A Α A2 A2 b . 43 A5 A4 As A3 A6 B3 B2 B1 Α B B2 B3 A2 C ...
... triangle are parts of diagonals А , and AA of the n - gon ; our triangle is one of the four triangles into which the quadrilateral А12 , is divided by its diagonals . Thus each A4 A Α A2 A2 b . 43 A5 A4 As A3 A6 B3 B2 B1 Α B B2 B3 A2 C ...
第 187 頁
... triangles with vertices at these points has A , as a vertex . It is clear that the second vertex A of this triangle ( that is , second with respect to the order A1 , A2 , ... , Aзn ) must be one of the points A2 , A5 , A8 , ... , A3k ...
... triangles with vertices at these points has A , as a vertex . It is clear that the second vertex A of this triangle ( that is , second with respect to the order A1 , A2 , ... , Aзn ) must be one of the points A2 , A5 , A8 , ... , A3k ...
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A₁ A₂ An+m arrangements b₁ b₂ binomial coefficients binomial theorem bishops black squares 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 length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain P₁ 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