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 23 筆
第 142 頁
... Combining each of the nine possible values for the first digit with each of the nine possible values for the second digit , each of the nine possible values for the third , and each of the nine possible values for the fourth , we obtain ...
... Combining each of the nine possible values for the first digit with each of the nine possible values for the second digit , each of the nine possible values for the third , and each of the nine possible values for the fourth , we obtain ...
第 145 頁
... Combining each of the 12 possibilities for the first person with each of the 12 possibilities for the second person , we obtain 12.12 = 12a equally likely possible outcomes for the first two people . Similarly we will have 123 equally ...
... Combining each of the 12 possibilities for the first person with each of the 12 possibilities for the second person , we obtain 12.12 = 12a equally likely possible outcomes for the first two people . Similarly we will have 123 equally ...
第 157 頁
... combining each of the 2 ) corresponding possible outcomes for the drawing ( 2n ) ( 2n 2 possibilities 2 ( 2n 2 4 ) possible thus obtained for the first two pairs of balls with the outcomes for the drawing of the third pair of balls ...
... combining each of the 2 ) corresponding possible outcomes for the drawing ( 2n ) ( 2n 2 possibilities 2 ( 2n 2 4 ) possible thus obtained for the first two pairs of balls with the outcomes for the drawing of the third pair of balls ...
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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 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