Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
搜尋書籍內容
第 1 到 3 筆結果,共 32 筆
第 48 頁
... remaining colors , say red , and turn the cube so that the red face is the back face . This can always be done by rotating the cube about the vertical axis through its center ( which leaves the blue face on top and the green face on the ...
... remaining colors , say red , and turn the cube so that the red face is the back face . This can always be done by rotating the cube about the vertical axis through its center ( which leaves the blue face on top and the green face on the ...
第 61 頁
... remaining 17. The x2 can be selected from any of the 20 factors ( 1 + x5 + x ) . Let us discuss for definiteness the case in which it is selected from the first factor . Then there are 19 remaining factors . From two of these we must ...
... remaining 17. The x2 can be selected from any of the 20 factors ( 1 + x5 + x ) . Let us discuss for definiteness the case in which it is selected from the first factor . Then there are 19 remaining factors . From two of these we must ...
第 142 頁
... remaining cards can be drawn ) ; corresponding to each of these outcomes there are four possible outcomes to the ... remaining D cards on the third draw , and the one remaining O card on the fourth draw . Combining each of the three ...
... remaining cards can be drawn ) ; corresponding to each of these outcomes there are four possible outcomes to the ... remaining D cards on the third draw , and the one remaining O card on the fourth draw . Combining each of the three ...
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 length mathematical induction maximum number multiple 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