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 23 筆
第 89 頁
... pieces 2 squares wide and 2 squares high . From the fact that none of these pieces can contain more than one king , it follows that the required maximum number of kings is at most k2 . But it is possible to arrange k2 kings on a 2k × 2k ...
... pieces 2 squares wide and 2 squares high . From the fact that none of these pieces can contain more than one king , it follows that the required maximum number of kings is at most k2 . But it is possible to arrange k2 kings on a 2k × 2k ...
第 102 頁
... pieces by the other lines and consequently increases the total number of pieces by 2n + 1. It follows from this that the total number of pieces is ( n + 1 ) 2 + n ( 2n + 1 ) = 3n2 + 3n + 1 . 44a . It is clear that n lines will divide ...
... pieces by the other lines and consequently increases the total number of pieces by 2n + 1. It follows from this that the total number of pieces is ( n + 1 ) 2 + n ( 2n + 1 ) = 3n2 + 3n + 1 . 44a . It is clear that n lines will divide ...
第 103 頁
... pieces , drawing the ( k + 1 ) st line increases the number of pieces by k + 1. But if only one line is drawn , it will divide the plane into two pieces . It follows from this B Fig . 49 that after n lines have been drawn the plane will ...
... pieces , drawing the ( k + 1 ) st line increases the number of pieces by k + 1. But if only one line is drawn , it will divide the plane into two pieces . It follows from this B Fig . 49 that after n lines have been drawn the plane will ...
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常見字詞
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