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 33 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 ...
... 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 ...
第 114 頁
... triangle are parts of diagonals А1 , 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 a . A2 A B1 B 85 B2 B3 BA A3 4 A2 b . A5 A3 A4 A6 B ...
... triangle are parts of diagonals А1 , 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 a . A2 A B1 B 85 B2 B3 BA A3 4 A2 b . A5 A3 A4 A6 B ...
第 117 頁
... triangle has three sides ; consequently , the total number of sides of all the triangles in such a decomposition is 3 · ( n − 2 ) . But each diagonal involved is a side of two of the triangles , and each side of the n - gon is a side ...
... triangle has three sides ; consequently , the total number of sides of all the triangles in such a decomposition is 3 · ( n − 2 ) . But each diagonal involved is a side of two of the triangles , and each side of the n - gon is a side ...
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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