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 February 26th, 2018, 04:26 PM #1 Member   Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0 Discrete Math question Any ideas how to do this? Thanks Last edited by sita; February 26th, 2018 at 04:29 PM.
 February 26th, 2018, 04:58 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1156 I don't know about the hint, but $\dbinom{2n}{2} = \dfrac{(2n)!}{2!(2n-2)!} = \dfrac{(2n)(2n-1)}{2} = n(2n-1)$ $2\dbinom{n}{2} = 2\dfrac{n!}{2(n-2)!} =n(n-1)$ $\dbinom{2n}{2}-2\dbinom{n}{2} = 2n^2 - n - (n^2-n) = (2n^2-n^2)-(n-n) = n^2$ $n^2 + 2 \dbinom{n}{2} = \dbinom{2n}{2}$
 February 26th, 2018, 10:06 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1156 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, 04:06 AM   #4
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Quote:
 Originally Posted by sita Any ideas how to do this? Thanks
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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