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 12 筆
第 58 頁
... factors , two of the factors are equal ( and the third factor different from them ) , then the factorization occurs three times : the unduplicated factor can come either first , second , or third . -- Let us compute the number of such ...
... factors , two of the factors are equal ( and the third factor different from them ) , then the factorization occurs three times : the unduplicated factor can come either first , second , or third . -- Let us compute the number of such ...
第 60 頁
... factors of the form a2 ... · ακ N = P11p22 Pkk , 7 ( N ) = ( a1 + 1 ) ( a2 + 1 ) · · · ( ax + 1 ) , 2 then and σ ( N ) Pi - 1 P2 - 1 på1 + 1 - 1 pa2 + 1 - 1 +1 Pk - ― 1 = • 19. The factorization of 126,000 into primes is 126,000 24 32 ...
... factors of the form a2 ... · ακ N = P11p22 Pkk , 7 ( N ) = ( a1 + 1 ) ( a2 + 1 ) · · · ( ax + 1 ) , 2 then and σ ( N ) Pi - 1 P2 - 1 på1 + 1 - 1 pa2 + 1 - 1 +1 Pk - ― 1 = • 19. The factorization of 126,000 into primes is 126,000 24 32 ...
第 61 頁
... factor on the right either 1 or x5 or x ' , and then multiplying the selected terms together . For example , if we took x5 from the first factor , x ' from the second factor , and 1 from all the other factors we would get the term x5 ...
... factor on the right either 1 or x5 or x ' , and then multiplying the selected terms together . For example , if we took x5 from the first factor , x ' from the second factor , and 1 from all the other factors we would get the term x5 ...
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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