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 30 筆
第 4 頁
... chosen from S in n ways . Once it is chosen , the second element can be chosen in n k -- 1 ways , etc. Hence P1 = n ( n − 1 ) ( n − 2 ) · · · ( n − k + 1 ) , - where there are k factors on the right . From ( 1 ) we now obtain Pr n ...
... chosen from S in n ways . Once it is chosen , the second element can be chosen in n k -- 1 ways , etc. Hence P1 = n ( n − 1 ) ( n − 2 ) · · · ( n − k + 1 ) , - where there are k factors on the right . From ( 1 ) we now obtain Pr n ...
第 4 頁
... chosen from S in n ways . Once it is chosen , the second element can be chosen in n - 1 ways , etc. Hence P1 = n ( n − 1 ) ( n − 2 ) · · · ( n − k + 1 ) , - - - where there are k factors on the right . From ( 1 ) we now obtain ...
... chosen from S in n ways . Once it is chosen , the second element can be chosen in n - 1 ways , etc. Hence P1 = n ( n − 1 ) ( n − 2 ) · · · ( n − k + 1 ) , - - - where there are k factors on the right . From ( 1 ) we now obtain ...
第 147 頁
... chosen in three ways and the passengers who sit there can be chosen in ( 2 ) ways . Then the car in which three passengers sit can be chosen in two ways , and the passengers who sit there can be chosen in ( 3 ) ways . The other ...
... chosen in three ways and the passengers who sit there can be chosen in ( 2 ) ways . Then the car in which three passengers sit can be chosen in two ways , and the passengers who sit there can be chosen in ( 3 ) ways . The other ...
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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