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swissgirl

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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  1. Monkeyball
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    Ma twin @Mimi_x3

    • one year ago
  2. NotTim
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    ottawa u. is that a university in those canada regions?

    • one year ago
  3. NotTim
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    btw, we can't access.

    • one year ago
  4. NotTim
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    unless you want us to hack.

    • one year ago
  5. swissgirl
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    ohhh s*** okkk give me a sec

    • one year ago
  6. NotTim
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    bad language pippa

    • one year ago
  7. swissgirl
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    alrighty here is the image

    • one year ago
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  8. NotTim
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    please some1 else be able to do this...

    • one year ago
  9. Valpey
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    \[\dbinom{x+a}{n}=\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{n-k}}\] \[\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{n-k}}=\dbinom{x}{0}\dbinom{a}{n-0}+\dbinom{x}{1}\dbinom{a}{n-1}+\dbinom{x}{2}\dbinom{a}{n-2}+...\] \[+\dbinom{x}{n}\dbinom{a}{0}\]

    • one year ago
  10. swissgirl
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    What rules did u use?

    • one year ago
  11. experimentX
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    easier method http://en.wikipedia.org/wiki/Vandermonde's_identity#Combinatorial_proof

    • one year ago
  12. Valpey
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    Win ^^

    • one year ago
  13. swissgirl
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    hahahahahah nahhhh both u guys win ok ill medal valpey and valpey medals experimentx

    • one year ago
  14. Valpey
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    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.

    • one year ago
  15. Valpey
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    But the proof using Democrats and Republicans in the US Senate works for me.

    • one year ago
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