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 14 筆
第 50 頁
... digits to ten . The new sequence of numbers consists of 1010 integers , beginning with 0,000,000,000 and ending with 9,999,999,999 . If one of these numbers has no 1's among its digits , then its first digit must be one of the nine ...
... digits to ten . The new sequence of numbers consists of 1010 integers , beginning with 0,000,000,000 and ending with 9,999,999,999 . If one of these numbers has no 1's among its digits , then its first digit must be one of the nine ...
第 51 頁
... digits ; the second ten - row consists entirely of numbers with 1's among their digits . Consequently , the first hundred - row contains 109.1 numbers which have 1's among their digits : ten in the second ten - row and one in each of ...
... digits ; the second ten - row consists entirely of numbers with 1's among their digits . Consequently , the first hundred - row contains 109.1 numbers which have 1's among their digits : ten in the second ten - row and one in each of ...
第 198 頁
... digits 12. Consequently , the probability that 2 " ends in the digits 12 is 1/20 = 0.05 Remark . It can be proved that the last k digits of the number 2 " are repeated in groups of 45 - 1 , starting with the number 2 ( see Shklyarskii ...
... digits 12. Consequently , the probability that 2 " ends in the digits 12 is 1/20 = 0.05 Remark . It can be proved that the last k digits of the number 2 " are repeated in groups of 45 - 1 , starting with the number 2 ( see Shklyarskii ...
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A₁ A₂ An+m arrangements b₁ b₂ binomial coefficients binomial theorem bishops black squares 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 n-gon number of different number of favorable number of paths number of shortest obtain P₁ 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