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 22 筆
第 173 頁
... segment A¡A + 1 is horizontal . But even such a point A , need not be illuminated , for the segment AA + 1 may lie in the shade cast by a later vertical segment . Denote by v the number of vertical unit segments of the path Ao , A1 ...
... segment A¡A + 1 is horizontal . But even such a point A , need not be illuminated , for the segment AA + 1 may lie in the shade cast by a later vertical segment . Denote by v the number of vertical unit segments of the path Ao , A1 ...
第 174 頁
... segments of AA , An + m in the shade . Thus , of the n points A , which had a chance of being illuminated , exactly m are eliminated by virtue of the shadow cast on the segment A¡A1 + 1 . Hence there remain nm illuminated points among ...
... segments of AA , An + m in the shade . Thus , of the n points A , which had a chance of being illuminated , exactly m are eliminated by virtue of the shadow cast on the segment A¡A1 + 1 . Hence there remain nm illuminated points among ...
第 217 頁
... segment represent the same point of the circle is immaterial ) . For the triangle ABC to be obtuse ( that is , for ... segment AA ' our problem assumes the following form : two points B and C are selected at random on the segment AA ...
... segment represent the same point of the circle is immaterial ) . For the triangle ABC to be obtuse ( that is , for ... segment AA ' our problem assumes the following form : two points B and C are selected at random on the segment AA ...
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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