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April 27th, 2019, 06:01 PM   #1
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Combinatorial proof

Hi guys any idea how I can give combinatorial proof to the given identities?

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 April 28th, 2019, 01:39 AM #2 Senior Member   Joined: Apr 2017 From: New York Posts: 165 Thanks: 6 I have solved the second one ( C(2n,n) one). I am still working on the first one.
 April 28th, 2019, 02:11 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,105 Thanks: 2324 How do you prove the second one?
 April 28th, 2019, 09:02 AM #4 Senior Member   Joined: Apr 2017 From: New York Posts: 165 Thanks: 6 I will tell you if you show me the first one
 April 28th, 2019, 09:03 AM #5 Senior Member   Joined: Apr 2017 From: New York Posts: 165 Thanks: 6 by the way there isn't a subject titled "discrete math" so I had to post it under abstract algebra. Last edited by Leonardox; April 28th, 2019 at 09:47 AM.
April 28th, 2019, 10:03 AM   #6
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Joined: Apr 2017
From: New York

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number two is proven like this:

right handside is the square of C(n.k). separate the factors. C(n,k) * C(n.k)
and from the combinational identity we know that C(n,k) can be rewritten as C(n, n-k)

now using the Theorem C(m+n, k) = Sigma...

OK leT me attach the solution that is faster.
I have attached two photos.
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 April 28th, 2019, 02:38 PM #7 Senior Member   Joined: Sep 2016 From: USA Posts: 683 Thanks: 455 Math Focus: Dynamical systems, analytic function theory, numerics For the first one you should notice that the left hand side looks an awful lot like a derivative. In particular, what happens if you differentiate $(1 + x)^n = \sum_{k=0}^n {{n}\choose{k}} x^k$ and evaluate it at $x = 1$? Thanks from Leonardox
 April 28th, 2019, 07:19 PM #8 Senior Member   Joined: Apr 2017 From: New York Posts: 165 Thanks: 6 hm that looks interesting. did not thought of that. I should try. Also the hint in the question is : "Think about picking a club and its president." what do you think?

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