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 137 頁
... relation to be proved follows immediately . Note that this relation can also be derived easily from the result of problem 57g . For since be rewritten in the form n n ( * ) = ( n = m ) , m ― the relation to be proved can ( nm ) + ( m ) ...
... relation to be proved follows immediately . Note that this relation can also be derived easily from the result of problem 57g . For since be rewritten in the form n n ( * ) = ( n = m ) , m ― the relation to be proved can ( nm ) + ( m ) ...
第 189 頁
... relation connecting Sn + 1 with Sn , Sn - 1 , Sn - 2 , ... , S. Let A1A2 A2nA2n + 1A2n + 2 be a convex ( 2n + 2 ) gon ( fig . 67 ) ; for any decomposition of the ( 2n + 2 ) gon into quadrilaterals , consider the quadrilateral which ...
... relation connecting Sn + 1 with Sn , Sn - 1 , Sn - 2 , ... , S. Let A1A2 A2nA2n + 1A2n + 2 be a convex ( 2n + 2 ) gon ( fig . 67 ) ; for any decomposition of the ( 2n + 2 ) gon into quadrilaterals , consider the quadrilateral which ...
第 192 頁
... relation ( see footnote , p . 121 ) into the set S as follows . Two sequences σ1 and σ2 , are called equivalent if they differ only by " cyclic permutation ” ; i.e. , if σ can be gotten by moving a block of letters from the beginning of ...
... relation ( see footnote , p . 121 ) into the set S as follows . Two sequences σ1 and σ2 , are called equivalent if they differ only by " cyclic permutation ” ; i.e. , if σ can be gotten by moving a block of letters from the beginning of ...
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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