一個卡特蘭等式的組合證明

Abstract

本文所探討的是卡特蘭等式以及開票一路領先的問題，並將其結果推廣到高維度的卡特蘭數。假設有甲、乙兩位候選人，其得票數分別為m及n票，且m≧n，我們若將開票過程建立在直角座標上，起點由(0,0)開始，將甲得一票記作向量(1,0)，乙得一票記作向量(0,1)，則由甲候選人一路領先的開票方法數，即為直線y = x以下的路徑總數。\n 在本文中，我們利用一種對射函數，將好路徑對應到標準楊氏圖表上的數字填法，再利用勾長公式算出方法數，藉此來得到好路徑的總數，作為卡特蘭等式的一種組合證明。文末也此方法推廣應用到多位候選人的開票一路領先方式，並得到高維度的卡特蘭等式：\nC_(m,n)=((mn¦(n,n,n,..,n)))/(∏_(k=1)^(m-1)▒((n+k)¦k) )

In this thesis, we study the Catalan identity and generalize the results to obtain the higher dimensional Catalan identity. Suppose that there are two candidates A and B for an election. A receives m votes and B receives n votes with m≧n. If we consider the ballot as a lattice path on coordinate system, starting from (0,0), where every vote for A is expressed as a vector (1,0) and votes for B are expressed as vectors (0,1). Then the number of ways that A leads all the way equals to the number of paths under the diagonal y=x.\n In this paper, we establish a bijection function that corresponds the good paths to the Young tableaux, and calculate the number of Young tableaux by hook formula. Finally, we generalize this method to calculate the higher dimensional Catalan identity:\nC_(m,n)=((mn¦(n,n,n,..,n)))/(∏_(k=1)^(m-1)▒((n+k)¦k) )