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 55 筆
第 78 頁
... consequently , the total number of bishops cannot be more than 15. But not all 15 diagonals can be occupied , since the first and last of them each consist of a single square , and a bishop on either of these two squares would control ...
... consequently , the total number of bishops cannot be more than 15. But not all 15 diagonals can be occupied , since the first and last of them each consist of a single square , and a bishop on either of these two squares would control ...
第 142 頁
... Consequently , the required probability is 1 / 360≈ 0.003 . 65b . Here again the experiment consists of drawing four cards in suc- cession from a set of six cards ; consequently there are 360 possible outcomes to the experiment . But ...
... Consequently , the required probability is 1 / 360≈ 0.003 . 65b . Here again the experiment consists of drawing four cards in suc- cession from a set of six cards ; consequently there are 360 possible outcomes to the experiment . But ...
第 158 頁
... consequently , also the required probability p1 ) equals zero : if the total number of white balls is odd , at least one of the people must draw one white ball and one black ball . Hence we have only to consider the case of even n = 2k ...
... consequently , also the required probability p1 ) equals zero : if the total number of white balls is odd , at least one of the people must draw one white ball and one black ball . Hence we have only to consider the case of even n = 2k ...
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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