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 24 筆
第 35 頁
... Finally , one can also interpret the term " at random " to mean that the probability that the midpoint of a chord lies within a given circle is proportional to the area of that circle . ( Note that the midpoint of a chord uniquely ...
... Finally , one can also interpret the term " at random " to mean that the probability that the midpoint of a chord lies within a given circle is proportional to the area of that circle . ( Note that the midpoint of a chord uniquely ...
第 76 頁
... Finally , there will be only one possibility left for the last rook once the locations of the first n 1 have been specified . Pairing each of the n different possibilities for the first rook with each of the corresponding n - 1 ...
... Finally , there will be only one possibility left for the last rook once the locations of the first n 1 have been specified . Pairing each of the n different possibilities for the first rook with each of the corresponding n - 1 ...
第 198 頁
... finally one 2 - digit number which starts with a 1 ( namely , 16 ) . Therefore , q ( N ) = n − 1 . - ― If n is the number of digits in the number 2 , then n 1 is the characteristic of the logarithm ( to the base 10 ) of 2 , that is ...
... finally one 2 - digit number which starts with a 1 ( namely , 16 ) . Therefore , q ( N ) = n − 1 . - ― If n is the number of digits in the number 2 , then n 1 is the characteristic of the logarithm ( to the base 10 ) of 2 , that is ...
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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