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 27 筆
第 41 頁
... passing through A , B , C ( fig . 16 ) . Moreover t does not pass through D , by hypothesis . To be equidistant from the four points , s must be concentric with or parallel to t ( according as t is a circle or a straight line ) and must ...
... passing through A , B , C ( fig . 16 ) . Moreover t does not pass through D , by hypothesis . To be equidistant from the four points , s must be concentric with or parallel to t ( according as t is a circle or a straight line ) and must ...
第 139 頁
... pass through the point ( m - k + 1 , k − 1 ) ; k m - - is the number of shortest paths from the point ( 0,0 ) to the point ( n + m − k , k ) which pass through the point ( m - k + 2 , k − 2 ) , etc. and 2 ) ( 2 ) ( m ) ( 1 ) k is the ...
... pass through the point ( m - k + 1 , k − 1 ) ; k m - - is the number of shortest paths from the point ( 0,0 ) to the point ( n + m − k , k ) which pass through the point ( m - k + 2 , k − 2 ) , etc. and 2 ) ( 2 ) ( m ) ( 1 ) k is the ...
第 181 頁
... pass through the point D1 is equal to NAD1ND1An + m = 2N40'D1 ND1An + m ' and the number of shortest paths from A to An + m which pass through the point D1 is NAD , ND14n + m Consequently , there are exactly twice as many of the first ...
... pass through the point D1 is equal to NAD1ND1An + m = 2N40'D1 ND1An + m ' and the number of shortest paths from A to An + m which pass through the point D1 is NAD , ND14n + m Consequently , there are exactly twice as many of the first ...
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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