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 37 筆
第 76 頁
... arrangements of n rooks satisfy our conditions . Let us call the rook in the first column the first rook , that in ... arrangements of the first two rooks ; arguing similarly , there will be n ( n - 1 ) ( n - 2 ) possible arrangements of ...
... arrangements of n rooks satisfy our conditions . Let us call the rook in the first column the first rook , that in ... arrangements of the first two rooks ; arguing similarly , there will be n ( n - 1 ) ( n - 2 ) possible arrangements of ...
第 172 頁
... arrangement is finally recovered , we obtain a total of n + m - 1 new arrangements . Adding to these the original arrangement , we obtain a total of n + m arrangements ; we will now show that exactly n them are favorable , i.e. are such ...
... arrangement is finally recovered , we obtain a total of n + m - 1 new arrangements . Adding to these the original arrangement , we obtain a total of n + m arrangements ; we will now show that exactly n them are favorable , i.e. are such ...
第 174 頁
... arrangements of customers obtained from any one arrangement by successively putting the first customer at the end of the line have more people with five - dollar bills than with tens in front of each customer . — = Note that the n + m ...
... arrangements of customers obtained from any one arrangement by successively putting the first customer at the end of the line have more people with five - dollar bills than with tens in front of each customer . — = Note that the n + m ...
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