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February 26th, 2018, 03:26 PM  #1 
Newbie Joined: Oct 2017 From: Rumba Posts: 20 Thanks: 0  Discrete Math question Any ideas how to do this? Thanks Last edited by sita; February 26th, 2018 at 03:29 PM. 
February 26th, 2018, 03:58 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,098 Thanks: 1093 
I don't know about the hint, but $\dbinom{2n}{2} = \dfrac{(2n)!}{2!(2n2)!} = \dfrac{(2n)(2n1)}{2} = n(2n1)$ $2\dbinom{n}{2} = 2\dfrac{n!}{2(n2)!} =n(n1)$ $\dbinom{2n}{2}2\dbinom{n}{2} = 2n^2  n  (n^2n) = (2n^2n^2)(nn) = n^2$ $n^2 + 2 \dbinom{n}{2} = \dbinom{2n}{2}$ 
February 26th, 2018, 09:06 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,098 Thanks: 1093 
What the hint is referring to is the idea of splitting the $2n$ items into 2 mutually exclusive sets of $n$ each. Each of these two subsets has $\dbinom{n}{2}$ ways of selecting 2 items from them. This is the $2\dbinom{n}{2}$ term. Then you can pair up 1 item from each subset. These will be distinct from any of the pairs chosen entirely from one of the subsets. There are $n^2$ pairs you can make. The sum of these two terms account for all the pairs you can make from the original $2n$ items. 
February 27th, 2018, 03:06 AM  #4 
Senior Member Joined: Feb 2010 Posts: 688 Thanks: 131  Maybe a story something like this. In how many ways can you choose $\displaystyle 2$ people from a group of $\displaystyle n$ boys and $\displaystyle n$ girls? You can either choose one of each gender $\displaystyle n \times n = n^2$ or you can choose both the same gender $\displaystyle 2 \binom{n}{2}$ (multiply by two since there are two genders). 

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