swissgirl
  • swissgirl
Give a combinatorial proof of Vandemonde's identity, for x, a, n ∈ ℕ Look at image below where ( ⋅ ) denotes the binomial coefficient nCr.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Ma twin @Mimi_x3
NotTim
  • NotTim
ottawa u. is that a university in those canada regions?
NotTim
  • NotTim
btw, we can't access.

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NotTim
  • NotTim
unless you want us to hack.
swissgirl
  • swissgirl
ohhh s*** okkk give me a sec
NotTim
  • NotTim
bad language pippa
swissgirl
  • swissgirl
alrighty here is the image
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NotTim
  • NotTim
please some1 else be able to do this...
Valpey
  • Valpey
\[\dbinom{x+a}{n}=\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{n-k}}\] \[\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{n-k}}=\dbinom{x}{0}\dbinom{a}{n-0}+\dbinom{x}{1}\dbinom{a}{n-1}+\dbinom{x}{2}\dbinom{a}{n-2}+...\] \[+\dbinom{x}{n}\dbinom{a}{0}\]
swissgirl
  • swissgirl
What rules did u use?
experimentX
  • experimentX
easier method http://en.wikipedia.org/wiki/Vandermonde's_identity#Combinatorial_proof
Valpey
  • Valpey
Win ^^
swissgirl
  • swissgirl
hahahahahah nahhhh both u guys win ok ill medal valpey and valpey medals experimentx
Valpey
  • Valpey
The tricky part is the leap from: \[\left(\sum_{i=0}^{m}\dbinom{m}{i}x^i\right)\left(\sum_{j=0}^{n}\dbinom{n}{j}x^j\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}\dbinom{m}{k}\dbinom{n}{r-k}\right)x^r\] 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.
Valpey
  • Valpey
But the proof using Democrats and Republicans in the US Senate works for me.

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