English  |  正體中文  |  简体中文  |  Post-Print筆數 : 27 |  Items with full text/Total items : 112721/143689 (78%)
Visitors : 49588554      Online Users : 686
RC Version 6.0 © Powered By DSPACE, MIT. Enhanced by NTU Library IR team.
Scope Tips:
  • please add "double quotation mark" for query phrases to get precise results
  • please goto advance search for comprehansive author search
  • Adv. Search
    HomeLoginUploadHelpAboutAdminister Goto mobile version
    政大機構典藏 > 理學院 > 應用數學系 > 學位論文 >  Item 140.119/131112
    Please use this identifier to cite or link to this item: https://nccur.lib.nccu.edu.tw/handle/140.119/131112


    Title: 關於一個組合等式的對射證明
    A Bijection Proof About a Combinatorial Identity
    Authors: 黃大維
    HUANG, TA-WEI
    Contributors: 李陽明
    Chen, Young-Ming
    黃大維
    HUANG, TA-WEI
    Keywords: 組合等式
    對射證明
    排列組合
    Combinatorial identity
    Bijection proof
    Combinatorics
    Date: 2020
    Issue Date: 2020-08-03 17:58:35 (UTC+8)
    Abstract: 本篇論文的主旨是要證明兩個排列組合n*C(2r,(n+r-1))*C(r,2r)與C(r,n)*(n-r)*C(r,(n+r-1))是否相
    等,並嘗試找出這兩個排列組合之間的關係與意義。
    本篇論文提供兩個證明此組合等式n*C(2r,(n+r-1))*C(r,2r)=C(r,n)*(n-r)*C(r,(n+r-1))的方法,第一個方法簡潔快速,但是只能單純證明等式的相等,無法看出左式與右式的關係。
    第二個方法是將左式與右式分別建構成一個集合,並在兩個集合之間建構
    一個函數。此函數的特性為一對一(one to one)且映成(onto),也就是對射函數(bijection function),利用此方法完成第二個方法的證明。
    The main topic in this paper is to prove the equality of these two expressions n*C(2r,(n+r-1))*C(r,2r)and C(r,n)*(n-r)*C(r,(n+r-1)).And attempt to find the relation between these two expressions and their meaning in combinatorics.
    In this paper, we provide two kinds of methods to solve the combinatorial identity n*C(2r,(n+r-1))*C(r,2r)=C(r,n)*(n-r)*C(r,(n+r-1)).The first way is simply proving the combinatorial identity by expansing the formula without knowing the meaning of the equation.
    The second method to solve the equation is constructing two sets consisting of the numbers of elements in n*C(2r,(n+r-1))*C(r,2r) and C(r,n)*(n-r)*C(r,(n+r-1)), respectively.And construct a bijective function to complete the second method of proof.
    Reference: [1] Alan Tucker(2012),Applied Combinatorics,sixth edition,John Wiley & Sons,Inc.,p.233.
    [2]C.L.Liu(2000),Introduction to Combinatorial Mathematics(International editions),McGraw-Hill.
    [3]Susanna S.Epp(2003),Discrete Mathematics with Applications,Cengage Learning
    [4]Peter J.Cameron(1994),Combinations:Topics,Techniques,Algorithms,London,
    Queen Mary&Westfield College.
    [5]劉麗珍(1994),一個組合等式的一對一證明,國立政治大學,應用數學碩士論文。
    [6]陳建霖(1996),一個組合等式的證明,國立政治大學,應用數學碩士論文。
    [7]韓淑惠(2011),開票一路領先的對射證明,國立政治大學,應用數學碩士論文。
    [8]薛麗姿(2013),一個珠狀排列的公式,國立政治大學,應用數學碩士論文。
    [9]黃永昌(2018),一個組合等式的對射證明,國立政治大學,應用數學碩士論文。
    Description: 碩士
    國立政治大學
    應用數學系
    108751008
    Source URI: http://thesis.lib.nccu.edu.tw/record/#G0108751008
    Data Type: thesis
    DOI: 10.6814/NCCU202000717
    Appears in Collections:[應用數學系] 學位論文

    Files in This Item:

    File Description SizeFormat
    100801.pdf837KbAdobe PDF20View/Open


    All items in 政大典藏 are protected by copyright, with all rights reserved.


    社群 sharing

    著作權政策宣告 Copyright Announcement
    1.本網站之數位內容為國立政治大學所收錄之機構典藏,無償提供學術研究與公眾教育等公益性使用,惟仍請適度,合理使用本網站之內容,以尊重著作權人之權益。商業上之利用,則請先取得著作權人之授權。
    The digital content of this website is part of National Chengchi University Institutional Repository. It provides free access to academic research and public education for non-commercial use. Please utilize it in a proper and reasonable manner and respect the rights of copyright owners. For commercial use, please obtain authorization from the copyright owner in advance.

    2.本網站之製作,已盡力防止侵害著作權人之權益,如仍發現本網站之數位內容有侵害著作權人權益情事者,請權利人通知本網站維護人員(nccur@nccu.edu.tw),維護人員將立即採取移除該數位著作等補救措施。
    NCCU Institutional Repository is made to protect the interests of copyright owners. If you believe that any material on the website infringes copyright, please contact our staff(nccur@nccu.edu.tw). We will remove the work from the repository and investigate your claim.
    DSpace Software Copyright © 2002-2004  MIT &  Hewlett-Packard  /   Enhanced by   NTU Library IR team Copyright ©   - Feedback