關於一個組合等式的對射證明

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\n等，並嘗試找出這兩個排列組合之間的關係與意義。\n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。\n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\n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.\nIn 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.\nThe 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.