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 8 筆
第 14 頁
... Problem 54 will reoccur later in another connection ( see problem 84a ) . At that point some related problems ( 84b and 84c ) will be given ; for more general results , see the remark at the end of the solution of problem 84c . 55a . A ...
... Problem 54 will reoccur later in another connection ( see problem 84a ) . At that point some related problems ( 84b and 84c ) will be given ; for more general results , see the remark at the end of the solution of problem 84c . 55a . A ...
第 185 頁
... problem 83a ( with m = n ) , the number of arrangements of 2n customers ( or points ) which ( n + 1 ) , we have satisfy our hypothesis is ( 2 ) ( 2n ) / ( n Fn = 1 2n n + 1 n We have thus obtained a new solution to problem 54 . Another ...
... problem 83a ( with m = n ) , the number of arrangements of 2n customers ( or points ) which ( n + 1 ) , we have satisfy our hypothesis is ( 2 ) ( 2n ) / ( n Fn = 1 2n n + 1 n We have thus obtained a new solution to problem 54 . Another ...
第 230 頁
... problem 83a , the answer to problem 54 is obtained immediately from this , Inasmuch as the answer to problem 53b can be derived from the answer to problem 54 ( compare with the solution to problem 54 ) , the reasoning indicated here also ...
... problem 83a , the answer to problem 54 is obtained immediately from this , Inasmuch as the answer to problem 53b can be derived from the answer to problem 54 ( compare with the solution to problem 54 ) , the reasoning indicated here also ...
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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