長度為n的三元數列(0, 1, 2)，探討(一)0為偶數個1為偶數個，或(二)0為偶數個1為奇數個，或(三)0為奇數個1為偶數個，或(四)0為奇數個1為奇數個的方法數時，就離散的傳統上來說是用遞迴關係去求解。本文將建構一對一函數，利用一對一函數的特性去求此問題的解，與以前的方法比較起來僅需要了解一對一函數的特性即可求解，易懂且不需要用到比較複雜的遞迴觀念。 / The problem of the number of ternary sequences of length n with :
(a) 0 is even, 1 is even,
(b) 0 is even, 1 is odd,
(c) 0 is odd, 1 is even,
(d) 0 is odd, 1 is odd,
has been solved by recurrence relations before. We solve the problem by constructingone-to-one functions, and use the characteristics of one-to-one functions to solve this problem. Our method is simpler than those methods which have been done before.
| Identifer | oai:union.ndltd.org:CHENGCHI/G0097751011 |
| Creators | 林宥廷 |
| Publisher | 國立政治大學 |
| Source Sets | National Chengchi University Libraries |
| Language | 中文 |
| Detected Language | English |
| Type | text |
| Rights | Copyright © nccu library on behalf of the copyright holders |
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