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April 27th, 2019, 06:01 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 165 Thanks: 6  Combinatorial proof
Hi guys any idea how I can give combinatorial proof to the given identities? Thanks in advance. 
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 
Senior Member Joined: Apr 2017 From: New York Posts: 165 Thanks: 6 
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, nk) now using the Theorem C(m+n, k) = Sigma... OK leT me attach the solution that is faster. I have attached two photos. 
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 $x = 1$? 
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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