二項式做法

因為 4C(n,4) + 5C(n,5) = n*C(n,4), 3C(n,3) + 4C(n,4) = n*C(n,3),
2C(n,2) + 3C(n,3) = n*C(n,2) 及 C(n,r) + C(n,r+1) = C(n+1,r+1),
所以
2C(n,2) + 9C(n,3) + 12C(n,4) + 5C(n,5)
= 2C(n,2) + 3C(n,3) + 3C(n,3) + 3C(n,3) + 4C(n,4) + 4C(n,4) + 4C(n,4) + 5C(n,5)
= n*C(n,2) + n*C(n,3) + n*C(n,3) + n*C(n,4)
= n*C(n+1,3) + n*C(n+1,4)
= n*C(n+2,4)

2C(n，2)+9C(n，3)+12C(n，4)+5C(n，5)
= 2(C(n,2)+C(n,3))+7(C(n,3)+C(n,4))+5(C(n,4)+C(n,5))
= 2C(n+1,3)+7C(n+1,4)+5C(n+1,5)
= 2C(n+2,4)+5C(n+2,5)
= 2C(n+3,5)+3C(n+2,5)
= [2(n+3)(n+2)(n+1)n(n-1)+3(n+2)(n+1)n(n-1)(n-2)]/5!
= [2(n+3)+3(n-2)](n+2)(n+1)n(n-1)/5!
= n^2(n+1)(n^2-1)/24