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 54 頁
... Reasoning as in part a we see that # ( C ) = [ 999 ] = = 333 , # ( ANC ) = [ 999 ] = 66 , 999 # ( BNC ) = and # ( AB O C ) = 21 9 . 105 . 15 = 999 By the principle of inclusion and exclusion , # ( AU BUC ) = 199 + 142 + 333 28 66 — 47 + ...
... Reasoning as in part a we see that # ( C ) = [ 999 ] = = 333 , # ( ANC ) = [ 999 ] = 66 , 999 # ( BNC ) = and # ( AB O C ) = 21 9 . 105 . 15 = 999 By the principle of inclusion and exclusion , # ( AU BUC ) = 199 + 142 + 333 28 66 — 47 + ...
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 ...
第 194 頁
... reasoning is reversible . As I continues moving to the left it strikes another pair of vertices A ,, Ar + m + n . The path P , corresponds to an outcome in S2 , for ( reasoning as above ) it contains one of the vertices A , A and one of ...
... reasoning is reversible . As I continues moving to the left it strikes another pair of vertices A ,, Ar + m + n . The path P , corresponds to an outcome in S2 , for ( reasoning as above ) it contains one of the vertices A , A and one 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