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 16 筆
第 50 頁
... marked ( the last of them will be 986 ) . The first number to be marked on the fourth time around would be 986 +15 — 1000 : = 1. Since this number has already been marked earlier , by continuing to count in 15's around the circle , we ...
... marked ( the last of them will be 986 ) . The first number to be marked on the fourth time around would be 986 +15 — 1000 : = 1. Since this number has already been marked earlier , by continuing to count in 15's around the circle , we ...
第 82 頁
... marked by a circle in Fig . 33 fig . 33 ) . Draw the two diagonals which pass through this square ; these diagonals end at two other outer squares , which are marked by crosses in fig . 33. Now draw the other diagonals through the points ...
... marked by a circle in Fig . 33 fig . 33 ) . Draw the two diagonals which pass through this square ; these diagonals end at two other outer squares , which are marked by crosses in fig . 33. Now draw the other diagonals through the points ...
第 99 頁
... marked by circles in fig . 47a ) . In this case we must leave empty the squares of the first rectangle which are marked by crosses : the two squares in the third row are controlled by the two knights , and the square in the second row ...
... marked by circles in fig . 47a ) . In this case we must leave empty the squares of the first rectangle which are marked by crosses : the two squares in the third row are controlled by the two knights , and the square in the second row ...
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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