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 29 筆
第 58 頁
... satisfy a1 + A2 + A3 : 1+ 2+ 3+ 4+ 5+ 6+ 7 = 28 . = 6 is By the same argument there are exactly 28 systems of numbers b1 , b2 , by which satisfy b1 + b2 + b = 6. Since any combination of such a triple of numbers a1 , a , a , with a ...
... satisfy a1 + A2 + A3 : 1+ 2+ 3+ 4+ 5+ 6+ 7 = 28 . = 6 is By the same argument there are exactly 28 systems of numbers b1 , b2 , by which satisfy b1 + b2 + b = 6. Since any combination of such a triple of numbers a1 , a , a , with a ...
第 92 頁
... satisfy this condition ; one such arrangement is shown in fig . 39 . It can be shown that on an ordinary chessboard there are 92 different arrangements of eight queens which satisfy the condition imposed . ( See , for example , M ...
... satisfy this condition ; one such arrangement is shown in fig . 39 . It can be shown that on an ordinary chessboard there are 92 different arrangements of eight queens which satisfy the condition imposed . ( See , for example , M ...
第 116 頁
... satisfy , it follows that ji , j2 , ... , jk - 1 satisfy the inequalities - - = - - 1 ≤ jï < j1⁄2 < j3 < · · · < jx - 1 ≤ ( n − 3 ) — ( k − 2 ) — n − k − 1 . Conversely , if j1 , j2 , ... , jk - 1 are distinct integers between 1 ...
... satisfy , it follows that ji , j2 , ... , jk - 1 satisfy the inequalities - - = - - 1 ≤ jï < j1⁄2 < j3 < · · · < jx - 1 ≤ ( n − 3 ) — ( k − 2 ) — n − k − 1 . Conversely , if j1 , j2 , ... , jk - 1 are distinct integers between 1 ...
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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