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 11 筆
第 99 頁
... rectangle ( we will call this the first rectangle ) . Let us first try putting knights in each of the bottom two squares of this rectangle ( these squares are marked by circles in fig . 47a ) . In this case we must leave empty the ...
... rectangle ( we will call this the first rectangle ) . Let us first try putting knights in each of the bottom two squares of this rectangle ( these squares are marked by circles in fig . 47a ) . In this case we must leave empty the ...
第 100 頁
... rectangle of the upper row , thus making it impossible to put any knights in that rectangle . We therefore see that the bottom two squares of the first rectangle cannot simultaneously be occupied . If there were no knights in the bottom ...
... rectangle of the upper row , thus making it impossible to put any knights in that rectangle . We therefore see that the bottom two squares of the first rectangle cannot simultaneously be occupied . If there were no knights in the bottom ...
第 112 頁
... rectangle k squares wide and I squares high can be chosen in ( 9 k ) ( 9-1 ) different ways , and that there are a total of ( 9 k ) ( 9-1 ) such rectangles on the board . — Let us determine how many rectangles of given width k there are ...
... rectangle k squares wide and I squares high can be chosen in ( 9 k ) ( 9-1 ) different ways , and that there are a total of ( 9 k ) ( 9-1 ) such rectangles on the board . — Let us determine how many rectangles of given width k there are ...
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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