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 27 筆
第 66 頁
... equation x + y + z = n . ― - Observe first of all that the equation y + z = k ( where k is a positive integer ) has k 1 positive integral solutions . In fact , in this case y can take of the values 1 , 2 , ... , k any 1 ( y cannot take ...
... equation x + y + z = n . ― - Observe first of all that the equation y + z = k ( where k is a positive integer ) has k 1 positive integral solutions . In fact , in this case y can take of the values 1 , 2 , ... , k any 1 ( y cannot take ...
第 132 頁
... equations ( 1 ) , ( 2 ) , ( 3 ) , and ( 4 ) by 1 , i , -1 , and i respectively , and add . This gives n 4 { ( 3 ) + ... equation x3 = 1 ) , which have the values x3 = 1 , −1 + i√√3 and 2 -1- i√3 2 We will use the notation w = ( −1 ...
... equations ( 1 ) , ( 2 ) , ( 3 ) , and ( 4 ) by 1 , i , -1 , and i respectively , and add . This gives n 4 { ( 3 ) + ... equation x3 = 1 ) , which have the values x3 = 1 , −1 + i√√3 and 2 -1- i√3 2 We will use the notation w = ( −1 ...
第 164 頁
... Equation ( 2 ) resembles the formula p + 1 when r = 1 when r > 1 P ( P + 1 ) = ( ) + ( 2 ) r - for the binomial coefficients . Just as this formula can be used to construct Pascal's triangle , equation ( 2 ) can be used to construct a ...
... Equation ( 2 ) resembles the formula p + 1 when r = 1 when r > 1 P ( P + 1 ) = ( ) + ( 2 ) r - for the binomial coefficients . Just as this formula can be used to construct Pascal's triangle , equation ( 2 ) can be used to construct a ...
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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