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 47 筆
第 46 頁
... consists of four straight lines through P. 2 , 3 1 , 3 , 4 19 4 Now let П1 , П2 , П ̧ , П be the planes of the faces of the tetrahedron T. The above remarks show that the locus of all points equidistant from ПI1 , П , and П , consists ...
... consists of four straight lines through P. 2 , 3 1 , 3 , 4 19 4 Now let П1 , П2 , П ̧ , П be the planes of the faces of the tetrahedron T. The above remarks show that the locus of all points equidistant from ПI1 , П , and П , consists ...
第 51 頁
... consists of the numbers from 10 to 19 ) will contain exactly one number with a 1 among its digits ; the second ten - row consists entirely of numbers with 1's among their digits . Consequently , the first hundred - row contains 10 + 9.1 ...
... consists of the numbers from 10 to 19 ) will contain exactly one number with a 1 among its digits ; the second ten - row consists entirely of numbers with 1's among their digits . Consequently , the first hundred - row contains 10 + 9.1 ...
第 53 頁
... consists of the numbers 5 , 10 , 15 , . . . , 995 – 199.5 . Similarly # ( B ) = [ 999/7 ] 142 , since B consists of the numbers 7 , 14 , 21 , ... , 994 = 142.7 . Now A OB consists of all positive integers ≤ 999 which are multiples of 5 ...
... consists of the numbers 5 , 10 , 15 , . . . , 995 – 199.5 . Similarly # ( B ) = [ 999/7 ] 142 , since B consists of the numbers 7 , 14 , 21 , ... , 994 = 142.7 . Now A OB consists of all positive integers ≤ 999 which are multiples of 5 ...
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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