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Give a combinatorial proof of Vandemonde's identity, for x, a, n ∈ ℕ
Look at image below
where ( ⋅ ) denotes the binomial coefficient nCr.
 one year ago
 one year ago
Give a combinatorial proof of Vandemonde's identity, for x, a, n ∈ ℕ Look at image below where ( ⋅ ) denotes the binomial coefficient nCr.
 one year ago
 one year ago

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NotTimBest ResponseYou've already chosen the best response.0
ottawa u. is that a university in those canada regions?
 one year ago

NotTimBest ResponseYou've already chosen the best response.0
unless you want us to hack.
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
ohhh s*** okkk give me a sec
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
alrighty here is the image
 one year ago

NotTimBest ResponseYou've already chosen the best response.0
please some1 else be able to do this...
 one year ago

ValpeyBest ResponseYou've already chosen the best response.2
\[\dbinom{x+a}{n}=\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{nk}}\] \[\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{nk}}=\dbinom{x}{0}\dbinom{a}{n0}+\dbinom{x}{1}\dbinom{a}{n1}+\dbinom{x}{2}\dbinom{a}{n2}+...\] \[+\dbinom{x}{n}\dbinom{a}{0}\]
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
What rules did u use?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
easier method http://en.wikipedia.org/wiki/Vandermonde's_identity#Combinatorial_proof
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
hahahahahah nahhhh both u guys win ok ill medal valpey and valpey medals experimentx
 one year ago

ValpeyBest ResponseYou've already chosen the best response.2
The tricky part is the leap from: \[\left(\sum_{i=0}^{m}\dbinom{m}{i}x^i\right)\left(\sum_{j=0}^{n}\dbinom{n}{j}x^j\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}\dbinom{m}{k}\dbinom{n}{rk}\right)x^r\] It is helpful to think of these terms as the diagonals of an m x n matrix of terms where each diagonal i+j=r.
 one year ago

ValpeyBest ResponseYou've already chosen the best response.2
But the proof using Democrats and Republicans in the US Senate works for me.
 one year ago
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