Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
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第 1 到 3 筆結果,共 25 筆
第 40 頁
... reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B or A ) on one side of it and the other three points on the other side of it . Thus there are a total of four planes in case 1 . 1 Consider case 2. Let A ...
... reasoning there is exactly one plane equidistant from A , B , C , D with C ( or B or A ) on one side of it and the other three points on the other side of it . Thus there are a total of four planes in case 1 . 1 Consider case 2. Let A ...
第 54 頁
... Reasoning as in part a we see that = # ( C ) = [ 999 ] = = 66 , = 333 , # ( A ~ C ) = [ 99 ] = 999 = · 9 . L105 # ( B ~ C ) = [ 1999 ] . 21 and # ( ABC ) : By the principle of inclusion and exclusion , # ( AUBUC ) 199 + 142 + 333286647 ...
... Reasoning as in part a we see that = # ( C ) = [ 999 ] = = 66 , = 333 , # ( A ~ C ) = [ 99 ] = 999 = · 9 . L105 # ( B ~ C ) = [ 1999 ] . 21 and # ( ABC ) : By the principle of inclusion and exclusion , # ( AUBUC ) 199 + 142 + 333286647 ...
第 88 頁
... reasoning used in parts a , b , c can be generalized to an n × n chessboard . In the following discussion we let ... Reasoning as in part a , we find that x = y = ( 2k ) ! ( 4k + 1 ) / 2 . Case 2 : n 4k + 1. Here there are 2k black ...
... reasoning used in parts a , b , c can be generalized to an n × n chessboard . In the following discussion we let ... Reasoning as in part a , we find that x = y = ( 2k ) ! ( 4k + 1 ) / 2 . Case 2 : n 4k + 1. Here there are 2k black ...
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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 points A1 polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose total number triangle unfavorable values vertex vertices