Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
搜尋書籍內容
第 1 到 3 筆結果,共 31 筆
第 54 頁
... Reasoning as in part a we see that = # ( C ) = [ 999 ] = 66 , = 333 , # ( A ~ c ) = [ 299 ] = 999 [ 999 # ( BC ) = and # ( AOB ( C ) = · 9 . 21 105 By the principle of inclusion and exclusion , # ( AU BUC ) 199 + 142 + 333 286647 +9 ...
... Reasoning as in part a we see that = # ( C ) = [ 999 ] = 66 , = 333 , # ( A ~ c ) = [ 299 ] = 999 [ 999 # ( BC ) = and # ( AOB ( C ) = · 9 . 21 105 By the principle of inclusion and exclusion , # ( AU BUC ) 199 + 142 + 333 286647 +9 ...
第 88 頁
... reasoning used in parts a , b , c can be generalized to an n x 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 x 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 Ar , Ar + m + n ° The path P , corresponds to an outcome in S2 , for ( reasoning as above ) it contains one of the vertices Ap , Ap + m + n ...
... reasoning is reversible . As I continues moving to the left it strikes another pair of vertices Ar , Ar + m + n ° The path P , corresponds to an outcome in S2 , for ( reasoning as above ) it contains one of the vertices Ap , Ap + m + n ...
其他版本 - 查看全部
常見字詞
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 given Hence inclusion and exclusion intersection k-gons knights L₁ 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 T₂ total number triangle unfavorable values vertex vertices