show that $ \sum_{k = 0}^N \frac{\binom{x}{k}\binom{N-x}{n-k}}{\binom{N}{n}} = \sum_{k = 0}^N \frac{\binom{n}{k}\binom{N-n}{x-k}}{\binom{N}{x}} $ I found this formula in a book
$$  \sum_{k = 0}^N \frac{\binom{x}{k}\binom{N-x}{n-k}}{\binom{N}{n}}
= \sum_{k = 0}^N \frac{\binom{n}{k}\binom{N-n}{x-k}}{\binom{N}{x}} $$
Notice $n \leftrightarrow x$.
Is there a proof using combinatorics?  What are both sides counting?  I know that $\binom{N}{k}$ counts the way of partitioning $N$ objects into groups of $k$ and $N-k$.  
I am really looking for a bijctive proof if possible
 A: Both sides equate to $1$. 
This comes from Hypergeometric Distribution.  For the LHS, the situation is that you have $N$ colored balls consisting of $x$ red balls and $N-x$ blue balls. 
Let $X$ be the number of red balls in a random selection of $n$ balls. Then we have
$$
P(X=k)=\frac{\binom xk \binom {N-x}{n-k}}{\binom Nn}.
$$
Summing this for $0\leq k \leq N$ (note that binomial coefficient $\binom ab$ is zero when $a<b$. ), we have
$$
1=\sum_{k=0}^N P(X=k) = \sum_{k=0}^N\frac{\binom xk \binom {N-x}{n-k}}{\binom Nn}.
$$
The RHS is treated similarly, and equates to $1$. Hence, both sides are $1$.
For the equality without the summation (which I think you meant in the first place, and @Dolma gave an algebraic proof), we prove that 
$$
\binom Nx\binom xk\binom{N-x}{n-k} = \binom Nn\binom nk\binom{N-n}{x-k}.
$$
The LHS is the number of ways to partition $\{1,\cdots ,N\}$ as a disjoint union $A\cup B\cup C\cup D$ where 
$$
|A|=k, \ |B|=x-k, \ |C|=n-k, \ |D|=N-x-n+k.
$$
On the other hand, the RHS is the number of ways to partition $\{1,\cdots ,N\}$ as a disjoint union $A\cup B\cup C\cup D$ as 
$$
|A|=k, \ |B|=n-k, \ |C|=x-k, \ |D|=N-x-n+k.
$$
Therefore, both sides are equal. 
A: The proof is quite immediate actually, all you need to do to prove it is simply to expand all the terms on each side of the equation.
If you restate your problem as:
$$
\sum_{k=0}^{N}F=\sum_{k=0}^{N}G
$$
Now just evaluate F and G:
$$
F=\frac{{x \choose k}{N-x \choose n-k}}{{N \choose n}}=\frac{\frac{x!}{k!\cdot (x-k)!}\cdot\frac{(N-x)!}{(n-k)!\cdot ((N-x)-(n-k))!}}{\frac{N!}{n!\cdot (N-n)!}}
$$
$$
\Rightarrow F=\frac{x!\cdot (N-x)!\cdot n!\cdot (N-n)!}{N!\cdot k!\cdot (x-k)!\cdot (n-k)!\cdot((N-x)-(n-k))!}
$$
Same thing for G gets you:
$$
G=\frac{{n \choose k}{N-n \choose x-k}}{{N \choose x}}=\frac{\frac{n!}{k!\cdot (n-k)!}\cdot\frac{(N-n)!}{(x-k)!\cdot ((N-n)-(x-k))!}}{\frac{N!}{x!\cdot (N-x)!}}
$$
$$
\Rightarrow G=\frac{n!\cdot (N-n)!\cdot x!\cdot (N-x)!}{N!\cdot k!\cdot (n-k)!\cdot (x-k)!\cdot((N-n)-(x-k))!}
$$
Now, let L be:
$$
L=\frac{x!\cdot (N-x)!\cdot n!\cdot (N-n)!}{N!\cdot k!\cdot (x-k)!\cdot (n-k)!}
$$
You can rewrite both terms F and G as:
$$
F=\frac{L}{((N-x)-(n-k))!}
$$
and
$$
G=\frac{L}{((N-n)-(x-k))!}
$$
Moreover:
$$
((N-n)-(x-k))! = (N-n-x+k)! = (N-x-n+k)! = ((N-x)-(n-k))!
$$
Thus, F=G and:
$$
\sum_{k=0}^{N}F=\sum_{k=0}^{N}G
$$
A: One can rearrange the identity and give a combinatorial proof, as i707107 did, but it’s really not much more than a disguised form of Vandermonde’s identity, which has a very easy combinatorial proof that you can find here (and in the answers to a number of questions on this site). By Vandermonde’s identity
$$\sum_{k=0}^N\binom{x}k\binom{N-x}{n-k}=\binom{N}n$$
and
$$\sum_{k=0}^N\binom{n}k\binom{N-n}{x-k}=\binom{N}x\;,$$
so both sides are $1$.
A: Let a set with $N$ elements be split up randomly twice. This in: 
two sets $A$ and $B$ that contain $n$ and $N-n$ elements respectively, and
two sets $U$ and $V$ that contain $x$ and $N-x$ respectively.
$|A\cap U|=k\wedge |A\cap V|=n-k\iff|A\cap U|=k\wedge |B\cap U|=x-k$ 
So these events have equal probabilities:
$$\frac{\binom{x}{k}\binom{N-x}{n-k}}{\binom{N}{n}}=\frac{\binom{n}{k}\binom{N-n}{x-k}}{\binom{N}{x}}$$
