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 38 筆
第 57 頁
... pairs ( x , y ) . Of these , 142 pairs are of the form ( x , x ) ; each other pair occurs twice , once as ( x , y ) and once as ( y , x ) . Consequently , the total number of different pairs ( x , y ) equals 1422 ―― 142 + 142 = 142 143 ...
... pairs ( x , y ) . Of these , 142 pairs are of the form ( x , x ) ; each other pair occurs twice , once as ( x , y ) and once as ( y , x ) . Consequently , the total number of different pairs ( x , y ) equals 1422 ―― 142 + 142 = 142 143 ...
第 60 頁
... pairs ( a1 , a ) with max ( a1 , a2 ) = 4 ; indeed there are 5 such pairs with a1 = 4 ( since a , can then be 0 , 1 , 2 , 3 , or 4 ) , and 5 pairs with a2 = 4 , but the pair ( 4,4 ) has been counted twice , so we get a total of 5 +51 9 ...
... pairs ( a1 , a ) with max ( a1 , a2 ) = 4 ; indeed there are 5 such pairs with a1 = 4 ( since a , can then be 0 , 1 , 2 , 3 , or 4 ) , and 5 pairs with a2 = 4 , but the pair ( 4,4 ) has been counted twice , so we get a total of 5 +51 9 ...
第 157 頁
... pair of balls with ( 2n 2 ) corresponding possible outcomes for the drawing of the second pair , then combining each of the ( 2n ) ( 22 ) possibilities 2 ( 2n 4 ) possible 2 thus obtained for the first two pairs of balls with the ...
... pair of balls with ( 2n 2 ) corresponding possible outcomes for the drawing of the second pair , then combining each of the ( 2n ) ( 22 ) possibilities 2 ( 2n 4 ) possible 2 thus obtained for the first two pairs of balls with 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