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 60 筆
第 61 頁
... Hence we have 4 possibilities involving one dime . Finally , if no dimes are used , we can use either 5 , 4 , 3 , 2 , 1 , or no nickels ( the remainder of the 25 ¢ consisting of pennies in each case ) . Hence we have 6 possibilities ...
... Hence we have 4 possibilities involving one dime . Finally , if no dimes are used , we can use either 5 , 4 , 3 , 2 , 1 , or no nickels ( the remainder of the 25 ¢ consisting of pennies in each case ) . Hence we have 6 possibilities ...
第 85 頁
... Hence there is one rook on each of these columns . Also , each of the 4 middle rows must contain a rook in order to control the 6 leftmost squares and the 6 right- most squares . Conversely , these conditions are sufficient to control ...
... Hence there is one rook on each of these columns . Also , each of the 4 middle rows must contain a rook in order to control the 6 leftmost squares and the 6 right- most squares . Conversely , these conditions are sufficient to control ...
第 87 頁
... Hence there are 3 · 10 · 2 = 60 solutions in this case . In case 2 , there are 3 ( 2 ) = 3 ways to pick the 2 columns which contain 2 rooks each . Once these are picked there are ( 2 ) = rooks on the first of the chosen columns . Then ...
... Hence there are 3 · 10 · 2 = 60 solutions in this case . In case 2 , there are 3 ( 2 ) = 3 ways to pick the 2 columns which contain 2 rooks each . Once these are picked there are ( 2 ) = rooks on the first of the chosen columns . Then ...
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