Part (b) first:
C(2N,2) = (2N)(2N1)/2 = 2NēN = N(N1) + Nē =2*N(N1)/2 + Nē = 2*C(N,2)+Nē
The left hand side says you have 2N objects and you choose 2 of them [C(2N,2)].
For the right hand side think of splitting these 2N objects into two equal groups. Your chosen object can either both be from the first group [C(N,2)], both be from the second group [C(N,2)] or be one object from each group [N*N].
The sum clearly works.
You can generalize this two any two group sizes.
If A+B=N
then C(N,2) = C(A,2)+C(B,2)+A*B
You can also generalize to more than 2 groups but it gets pretty messy.

Posted by Jer
on 20131101 08:18:40 