Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

swissgirl

  • 3 years ago

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

  • This Question is Closed
  1. Monkeyball
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ma twin @Mimi_x3

  2. NotTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ottawa u. is that a university in those canada regions?

  3. NotTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    btw, we can't access.

  4. NotTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    unless you want us to hack.

  5. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhh s*** okkk give me a sec

  6. NotTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    bad language pippa

  7. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alrighty here is the image

    1 Attachment
  8. NotTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please some1 else be able to do this...

  9. Valpey
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\dbinom{x+a}{n}=\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{n-k}}\] \[\sum_{k=0}^n{\dbinom{x}{k}\dbinom{a}{n-k}}=\dbinom{x}{0}\dbinom{a}{n-0}+\dbinom{x}{1}\dbinom{a}{n-1}+\dbinom{x}{2}\dbinom{a}{n-2}+...\] \[+\dbinom{x}{n}\dbinom{a}{0}\]

  10. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What rules did u use?

  11. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    easier method http://en.wikipedia.org/wiki/Vandermonde's_identity#Combinatorial_proof

  12. Valpey
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Win ^^

  13. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hahahahahah nahhhh both u guys win ok ill medal valpey and valpey medals experimentx

  14. Valpey
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The tricky part is the leap from: \[\left(\sum_{i=0}^{m}\dbinom{m}{i}x^i\right)\left(\sum_{j=0}^{n}\dbinom{n}{j}x^j\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}\dbinom{m}{k}\dbinom{n}{r-k}\right)x^r\] 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.

  15. Valpey
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    But the proof using Democrats and Republicans in the US Senate works for me.

  16. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy