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 10 筆
第 48 頁
... colors left , yellow , black , and white , with which to color the three remaining faces ( the front face and the two side faces ) . The front face can therefore be colored in three ways , each of which leaves two ways to color the side ...
... colors left , yellow , black , and white , with which to color the three remaining faces ( the front face and the two side faces ) . The front face can therefore be colored in three ways , each of which leaves two ways to color the side ...
第 101 頁
... color ; see fig . 47h ) , then the two remaining knights can only be put on the other two squares of the same color ; they are the only squares not controlled by the first two knights . Further , the only squares of the second rectangle ...
... color ; see fig . 47h ) , then the two remaining knights can only be put on the other two squares of the same color ; they are the only squares not controlled by the first two knights . Further , the only squares of the second rectangle ...
第 122 頁
... color , then C is equivalent only to itself . Therefore we have n equivalence classes ( one for each of the n colors ) consisting of a single coloring . If Co is a coloring in which at least two sectors have different colors , then we ...
... color , then C is equivalent only to itself . Therefore we have n equivalence classes ( one for each of the n colors ) consisting of a single coloring . If Co is a coloring in which at least two sectors have different colors , then we ...
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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