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 87 筆
第 83 頁
... obtain n - 2 different rectangles of the sort constructed above ( with vertices at the two crosses and the two circles ) . In each of these rectangles we must put a pair of bishops on opposite vertices . It is clear that for each rec ...
... obtain n - 2 different rectangles of the sort constructed above ( with vertices at the two crosses and the two circles ) . In each of these rectangles we must put a pair of bishops on opposite vertices . It is clear that for each rec ...
第 144 頁
... obtain a total of 8.6 when e If e = = 0 . = 5 , then a + c = 4 , b + d = = 48 solutions 9. Therefore ( a , c ) must be one of the pairs ( 0,4 ) , ( 4,0 ) , ( 1,3 ) , or ( 3,1 ) . In the first two cases , ( b , d ) can be any one of the ...
... obtain a total of 8.6 when e If e = = 0 . = 5 , then a + c = 4 , b + d = = 48 solutions 9. Therefore ( a , c ) must be one of the pairs ( 0,4 ) , ( 4,0 ) , ( 1,3 ) , or ( 3,1 ) . In the first two cases , ( b , d ) can be any one of the ...
第 145 頁
... obtained for the first two people with each of the corresponding 10 possibilities for the third person , etc. , we obtain a total of 12 11 · 10 · 1 12 ! favorable outcomes . sequently , the required probability is 12 ! / 1212≈ 0.000054 ...
... obtained for the first two people with each of the corresponding 10 possibilities for the third person , etc. , we obtain a total of 12 11 · 10 · 1 12 ! favorable outcomes . sequently , the required probability is 12 ! / 1212≈ 0.000054 ...
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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