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 29 筆
第 50 頁
... means that among the integers from 0,000,000,000 to 9,999,999,999 there are 910 different numbers which have no 1's ... mean a set of ten integers which consists of some multiple of ten and the next nine integers after it ( for example ...
... means that among the integers from 0,000,000,000 to 9,999,999,999 there are 910 different numbers which have no 1's ... mean a set of ten integers which consists of some multiple of ten and the next nine integers after it ( for example ...
第 124 頁
... means that if A。 is joined to some point A , in the polygon P ,, then A must be joined to Ak + t , A2k must be joined to A2 + t , etc. As in problem 55 we can show that the points Ao , Ax , A2x , ... , A ( p - 1 ) are all different ...
... means that if A。 is joined to some point A , in the polygon P ,, then A must be joined to Ak + t , A2k must be joined to A2 + t , etc. As in problem 55 we can show that the points Ao , Ax , A2x , ... , A ( p - 1 ) are all different ...
第 197 頁
... means that n ( n = divisible by 2.2.4.4 divisible by 128 ) . Thus , n ( n ― = 2 ) ( n - 6 ) is 4 ) ( n 64 ( it is easy to see that in this case it is even 1 ) ( n − 2 ) ( n − 3 ) ( n · = -4 ) ( n 5 ) ( n 6 ) is divisible by 64 if and ...
... means that n ( n = divisible by 2.2.4.4 divisible by 128 ) . Thus , n ( n ― = 2 ) ( n - 6 ) is 4 ) ( n 64 ( it is easy to see that in this case it is even 1 ) ( n − 2 ) ( n − 3 ) ( n · = -4 ) ( n 5 ) ( n 6 ) is divisible by 64 if and ...
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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