Question

Give a combinatorial proof of Vandemonde's identity, for x, a, n ∈ ℕ

Look at image below

where ( ⋅ ) denotes the binomial coefficient nCr.

Answer 1

Ma twin @Mimi_x3

Answer 2

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

Answer 3

btw, we can't access.

Answer 4

unless you want us to hack.

Answer 5

ohhh s*** okkk give me a sec

Answer 6

bad language pippa

Answer 7

alrighty here is the image

Answer 8

please some1 else be able to do this...

Answer 9

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

Answer 10

What rules did u use?

Answer 11

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

Answer 12

Win ^^

Answer 13

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

Answer 14

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.

Answer 15

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