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 33 筆
第 120 頁
... formulas is quite lengthy . However , by using Stirling's formula ? n ! ~ √2πn ( ) n one can obtain without difficulty an approximate formula for T1 : In - ( n ( n ( 2n = 4 ) ! ― - - - 1 ) [ ( n − 2 ) ! ] 2 22 ( n - 2 ) 1 ) √π ( n ...
... formulas is quite lengthy . However , by using Stirling's formula ? n ! ~ √2πn ( ) n one can obtain without difficulty an approximate formula for T1 : In - ( n ( n ( 2n = 4 ) ! ― - - - 1 ) [ ( n − 2 ) ! ] 2 22 ( n - 2 ) 1 ) √π ( n ...
第 127 頁
... formula for the sum of an infinite geometric pro- gression , 1 + x 1 + x3 = ( 1 + x ) 1 1 + = ( 1 + x ) ( 1 · x3 + x2 x2 + x12 - • ) . ( 1 ) ( This formula , of course , makes sense only if x has absolute value less than 1 , but we can ...
... formula for the sum of an infinite geometric pro- gression , 1 + x 1 + x3 = ( 1 + x ) 1 1 + = ( 1 + x ) ( 1 · x3 + x2 x2 + x12 - • ) . ( 1 ) ( This formula , of course , makes sense only if x has absolute value less than 1 , but we can ...
第 140 頁
... formula . ( 1 ) For n = 0 , ( 1 + x + x2 ) ° = 1 ; 0 that is , the formula holds ( since Bo 1 ) . Suppose now that this formula has already been established for the exponent n ; let us show that in this case it will also hold for the ...
... formula . ( 1 ) For n = 0 , ( 1 + x + x2 ) ° = 1 ; 0 that is , the formula holds ( since Bo 1 ) . Suppose now that this formula has already been established for the exponent n ; let us show that in this case it will also hold for the ...
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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