排列组合的证明A(n+1,n+1)-A(n,n)=n²A(n-1,n-1)
问题描述:
排列组合的证明A(n+1,n+1)-A(n,n)=n²A(n-1,n-1)
答
A(n,n) = n!
A(n+1,n+1) - A(n,n) = (n+1)!- n!
= (n+1)*n!- n!
= n*n!
= n*n*(n-1)!
= n^2A(n-1,n-1)