I'm asked to give a combinatorial proof of the following question, how would I go about doing it? 
The hint we're given is to think of it in terms of grid paths.
I don't necessarily want the answer, I'm interested in learning how to do this. Could anyone offer an explanation (clearly please) on how I would go about proving this?
 A: HINT: A grid path is a path in the grid $\Bbb N\times\Bbb N$, starting at $\langle 0,0\rangle$, in which the only allowed steps are from $\langle m,n\rangle$ to either $\langle m+1,n\rangle$ or $\langle m,n+1\rangle$. In other words, you can move one unit to the right or one unit up. The length of a path is the number of steps. 
A grid path to $\langle n,2n\rangle$ requires $3n$ steps, $n$ of them to the right and $2n$ up. These may be mixed in any order, so there are $\binom{3n}n$ ways to choose which $n$ steps will be to the right and therefore $\binom{3n}n$ grid paths to $\langle n,2n\rangle$.
Now let $p_k$ be the number of grid paths to $\langle n,2n\rangle$ that have exactly $k$ steps to the right in the first $2n$ steps; what is $p_k$? Remember that $\binom{n}k=\binom{n}{n-k}$.
A: Remember that $\binom{n}{k} = \binom{n}{n-k}$ and maybe you can figure it out.
A: This is probably not the proof that was intended, but many such binomial identities can be proven using generating functions.
If $p(x) = a_0 + a_1x + \ldots a_n x^n$ and $q(x) = b_0 + b_1x + \ldots b_n x^n$ are polynomials, recall how you find the coefficient of $x^n$ in the product $p(x)q(x)$:  you looking for all the pairs of terms $a_i x^i$, $b_j x^j$ where the exponents add up to $n$: $i+j = n$.  These terms multiply and contribute to the coefficient of $x^n$ in the product, so if you sum up over all such pairs you get the coefficient for $x^n$.  For example, the $x^2$ term in $$(1 + 2x + 4x^2)(2 + 7x + 5x^2 )$$ is $1\cdot 5x^2 + 2x \cdot 7x + 4x^2 \cdot 2$.  This leads to a general formula: the coefficient of $x^n$ in $p(x)q(x)$ is given by
$$\sum_{k=0}^n a_k b_{n-k}.$$ 
For this particular identity, you can use the binomial formula $$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$$
and let $p(x) = (1+x)^{2n}$ and $q(x) = (1+x)^n$.
