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 Two ways (Posted on 2013-10-31)
Show that if N is a positive integer, then C(2N,2)= 2*C(N,2)+ N^2

(a) using a combinatorial argument.
(b) by algebraic manipulation.

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 Solution | Comment 1 of 2
Part (b) first:
C(2N,2) = (2N)(2N-1)/2 = 2Nē-N = N(N-1) + Nē =2*N(N-1)/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 2013-11-01 08:18:40

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