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 57 筆
第 84 頁
... Hence there must be one rook on each of the 4 middle rows . Furthermore , there must be a rook on each of the 3 middle columns , for otherwise the 4 squares at the top and the 4 squares at the bottom would not be controlled . Con ...
... Hence there must be one rook on each of the 4 middle rows . Furthermore , there must be a rook on each of the 3 middle columns , for otherwise the 4 squares at the top and the 4 squares at the bottom would not be controlled . Con ...
第 85 頁
... Hence there is one rook on each of these columns . Also , each of the 4 middle rows must contain a rook in order to control the 6 leftmost squares and the 6 right- most squares . Conversely , these conditions are sufficient to control ...
... Hence there is one rook on each of these columns . Also , each of the 4 middle rows must contain a rook in order to control the 6 leftmost squares and the 6 right- most squares . Conversely , these conditions are sufficient to control ...
第 87 頁
... Hence there are 3 · 10 · 2 = 60 solutions in this case . 3 ways to pick the 2 columns which contain In case 2 , there are = 2 rooks each . Once these are picked there are ( 2 ) = 10 ways to put 2 rooks on the first of the chosen columns ...
... Hence there are 3 · 10 · 2 = 60 solutions in this case . 3 ways to pick the 2 columns which contain In case 2 , there are = 2 rooks each . Once these are picked there are ( 2 ) = 10 ways to put 2 rooks on the first of the chosen columns ...
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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