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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?

Thanks in advance.
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April 28th, 2019, 01:39 AM   #2
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I have solved the second one ( C(2n,n) one).
I am still working on the first one.
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April 28th, 2019, 02:11 AM   #3
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How do you prove the second one?
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April 28th, 2019, 09:02 AM   #4
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I will tell you if you show me the first one
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April 28th, 2019, 09:03 AM   #5
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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.
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April 28th, 2019, 10:03 AM   #6
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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
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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 $x = 1$?
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April 28th, 2019, 07:19 PM   #8
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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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