## 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.

1. Monkeyball

Ma twin @Mimi_x3

2. NotTim

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

3. NotTim

btw, we can't access.

4. NotTim

unless you want us to hack.

5. swissgirl

ohhh s*** okkk give me a sec

6. NotTim

7. swissgirl

alrighty here is the image

8. NotTim

please some1 else be able to do this...

9. Valpey

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

10. swissgirl

What rules did u use?

11. experimentX
12. Valpey

Win ^^

13. swissgirl

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

14. 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.

15. Valpey

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