關於組合數學的題目

1 Consider (1+x)^2n=(1+x)^n(1+x)^n
Expand both sides and the coefficient of x^n is
LHS: 2nCn
RHS: (nC0)(nCn)+(nC1)(nCn-1)+...+(nCn)(nC0)
=(nC0)(nC0)+(nC1)(nC1)+...+(nCn)(nCn)
=(nC0)^2+(nC1)^2+...+(nCn)^2
So
n
Σ C(n,i)^2 = C(2n,n) for all n屬於Z.
i=0

2 Using combinatorial argument.
Assume we have m+n objects and want to choose k objects, the combinations is just m+nCk
On the other hand, we can devide these m+n objects into two groups where one (Group A) has n objects and the other (Group B) has m objects. To choose k objects, we can choose 0 object from Group A and k objects from group B or 1 object from Group A and k-1 objects from group B or 2 objects from Group A and k-2 objects from group B, ect... The total combinations is (nC0)(mCk)+(nC1)(mCk-1)+...+(nCk)(mC0)
So
k
Σ C(n,i)*C(m,k-i) = C(m+n,k) where k<=min{m,n}
i=0

1. n至少要大於等於6吧，怎麼會對所有整數都成立
C(2,6)就沒意義了呀!?