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

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Monkeyball Group TitleBest ResponseYou've already chosen the best response.0
Ma twin @Mimi_x3
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
ottawa u. is that a university in those canada regions?
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
btw, we can't access.
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
unless you want us to hack.
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
ohhh s*** okkk give me a sec
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
bad language pippa
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
alrighty here is the image
 2 years ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
please some1 else be able to do this...
 2 years ago

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

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
What rules did u use?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
easier method http://en.wikipedia.org/wiki/Vandermonde's_identity#Combinatorial_proof
 2 years ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
hahahahahah nahhhh both u guys win ok ill medal valpey and valpey medals experimentx
 2 years ago

Valpey Group TitleBest 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.
 2 years ago

Valpey Group TitleBest ResponseYou've already chosen the best response.2
But the proof using Democrats and Republicans in the US Senate works for me.
 2 years ago
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