Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
搜尋書籍內容
第 1 到 3 筆結果,共 41 筆
第 152 頁
... third person draws the Hit slip in one - third of the cases ( that is , in four cases ) . Thus there are a total of 15 + 4 = 19 favorable outcomes . Hence the probability that at least one of the three hunters will hit the fox is 19/27 ...
... third person draws the Hit slip in one - third of the cases ( that is , in four cases ) . Thus there are a total of 15 + 4 = 19 favorable outcomes . Hence the probability that at least one of the three hunters will hit the fox is 19/27 ...
第 189 頁
... third vertex of this quadrilateral ( that is , third with respect to the order A1 , A2 , A3 , . . . , A2n + 2 ) must be one of the points A3 , A5 , A7 , . . . , A2n + 1 ( an even - numbered vertex of the ( 2n + 2 ) gon could not be the ...
... third vertex of this quadrilateral ( that is , third with respect to the order A1 , A2 , A3 , . . . , A2n + 2 ) must be one of the points A3 , A5 , A7 , . . . , A2n + 1 ( an even - numbered vertex of the ( 2n + 2 ) gon could not be the ...
第 229 頁
... Third solution . Consider the n + m arrangements which are obtained from a given arrangement by successively moving ... third solution of part a , the simplest way being the solution analogous to the third solution of part a . 84a . In a ...
... Third solution . Consider the n + m arrangements which are obtained from a given arrangement by successively moving ... third solution of part a , the simplest way being the solution analogous to the third solution of part a . 84a . In a ...
其他版本 - 查看全部
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