what is the meaning behind this combinatorial identity In the following comment: 
Solution of $\large\binom{x}{n}+\binom{y}{n}=\binom{z}{n}$ with $n\geq 3$
$$ \binom{2n-1}{n} + \binom{2n-1}{n} = \binom{2n}{n} $$
I'm wondering about the meaning of this equation - i.e why it is true.(of course I can see mathematically why this is true).
Also I'm thinking if there is any general equation that this is just private case (i.e equation that will right also for number which are not 2 or something like this)   
 A: HINT
$\large \binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k}$
basically all it says is that the coefficients in next row of pascal's triangle can be generated by adding the coefficients in current row :

Refer pascals rule
Also notice that $(2n-1) - (n-1) =  n$
A: The combinatorial explanation of 
$$\binom{N}{K} = \binom{N-1}{K} + \binom{N-1}{K-1}$$ can be obtained by
considering $\binom{N}{K}$ as the number of different subsets of cardinality $K$ that can be chosen from a set of $N$ distinct objects.  Instead of choosing
$K$ objects out of $N$, let us take a slightly different approach. Set aside
the $N$-th object and choose $K$ objects from the remaining $N-1$. Clearly,
we can form $\binom{N-1}{K}$ subsets this way (and none of them contain the
$N$-th object). Then, choose $K-1$ objects from the remaining $N-1$ objects
(we can do this in $\binom{N-1}{K-1}$ different ways) and add the $N$-th object
to make a subset of cardinality $K$.  All these subsets contain the $N$-th object.
Turning to your desired result, set $N = 2n$ and $K = n$ to get
$$\binom{2n}{n} = \binom{2n-1}{n} + \binom{2n-1}{n-1}$$ and observe
that $\binom{N}{K} = \binom{N}{N-K}$ (each subset of cardinaity $K$
corresponds to the complementary subset of cardinality $N-K$) implies that
$$\binom{2n-1}{n-1} = \binom{2n-1}{2n-1-(n-1)} = \binom{2n-1}{n}$$
and we are done.
