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 20 筆
第 137 頁
... paths into the following pairwise disjoint classes . The first class shall consist of those paths which start along the " horizontal " street ; there will be as many such paths as there are paths joining the intersection ( 1,0 ) to the ...
... paths into the following pairwise disjoint classes . The first class shall consist of those paths which start along the " horizontal " street ; there will be as many such paths as there are paths joining the intersection ( 1,0 ) to the ...
第 181 頁
... paths from A , to An + m which first meet L , at the point D , with the number of such shortest paths from Ao to An + m The number of shortest paths from Ao to An + m which pass through the point D1 is equal to NAD1ND1An + m = 2N40'D1 ...
... paths from A , to An + m which first meet L , at the point D , with the number of such shortest paths from Ao to An + m The number of shortest paths from Ao to An + m which pass through the point D1 is equal to NAD1ND1An + m = 2N40'D1 ...
第 182 頁
... paths from A , ' to An + m . Continuing to reason in the same way , we can prove that for any k the number of shortest paths from A to An + m which first meet the line Ĩ , at the point D is exactly twice the number of shortest paths ...
... paths from A , ' to An + m . Continuing to reason in the same way , we can prove that for any k the number of shortest paths from A to An + m which first meet the line Ĩ , at the point D is exactly twice the number of shortest paths ...
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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