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 103 頁
... drawn in the plane ; let us draw the ( k + 1 ) st line and see by how much it increases the number of pieces into which the plane is divided . The ( k + 1 ) st line meets each of the k lines which have already been drawn ; the k points ...
... drawn in the plane ; let us draw the ( k + 1 ) st line and see by how much it increases the number of pieces into which the plane is divided . The ( k + 1 ) st line meets each of the k lines which have already been drawn ; the k points ...
第 142 頁
... draw , the E on the second draw , the A on the third draw , and the F on the fourth draw . Consequently , the required probability is 1 / 360≈ 0.003 . 65b . Here again the experiment consists of drawing four cards in suc- cession from ...
... draw , the E on the second draw , the A on the third draw , and the F on the fourth draw . Consequently , the required probability is 1 / 360≈ 0.003 . 65b . Here again the experiment consists of drawing four cards in suc- cession from ...
第 156 頁
... draw any of - ( 2n ) = 2n ( 2n − 1 ) 2 2n second person can draw any of the - 2 2 pairs of balls . Then the = ( 2n - 2 ) ( 2n - 3 ) / 2 pairs which can be formed from the remaining 2n - 2 balls . The third can draw any of ― 5 ) / 2 ...
... draw any of - ( 2n ) = 2n ( 2n − 1 ) 2 2n second person can draw any of the - 2 2 pairs of balls . Then the = ( 2n - 2 ) ( 2n - 3 ) / 2 pairs which can be formed from the remaining 2n - 2 balls . The third can draw any of ― 5 ) / 2 ...
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 given Hence inclusion and exclusion intersection k-gons knights length mathematical induction maximum number multiple 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