（1）几个知识点：(!)C(m,n)=n!/[m!*(n-m)!].(!)(m+1)!=(m+1)*m!.(2)证明：因C(m,n)=n!/[m!*(n-m)!]=[(n+1-m)*n!]/[m!*(n+1-m)!].C(m-1,n)=n!/[(m-1)!*(n+1-m)!]=[m*n!]/[m!*(n+1-m)!].故右边=C(m,n)+C(m-1,n)={[(n+1-m)*n!]/[m!*(n+1-m)!]}+{[m*n!]/[m!*(n+1-m)!]}={n!*[(n+1-m)+m]}/[m!*(n+1-m)!]=(n+1)!/[m!*(n+1-m)!].显然,左边=C(m,n+1)=(n+1)!/[m!*(n+1-m)!].故左边=右边,即C(m,n+1)=C(m,n)+C(m-1,n).