I want to find out how the function: $$f(m)=\sum_{j=0}^m {n \choose j}{2n+m-j-1\choose m-j}(-2)^j$$ ($m$ is an integer and $n\gg m> 0$), scales with $n$, as $n\to\infty$. I am not good at combinatorics formulas, but from naive calculation of $f(m)$ for different values of $m$, I think it should scale like $n^{\lfloor{m/2} \rfloor }$.
(This function $f(m)$ is from the answer to my previous question)
As René Gy points out, $${n \choose j} = \frac{n(n-1)...(n-j+1)}{j!}$$
is a polynomial of degree $j$ in the variable $n$. As $n \to \infty$, a polynomial becomes dominated by its leading (highest degree) term, which eventually dwarfs all the other terms. The leading term in $n \choose j$ is $\frac 1{j!}n^j$.
Similarly, $${2n+m-j-1\choose m-j}= \frac{(2n+m-j-1)(2n+m-j-2)...(2n)}{(m-j)!}$$ is a polynomial of degree $m-j$, with leading term $\frac {2^{m-j}}{(m-j)!}n^{m-j}$
So ${n \choose j}{2n+m-j-1\choose m-j}(-2)^j$ will be a polynomial with leading term $$\left(\frac 1{j!}n^j\right)\left(\frac {2^{m-j}}{(m-j)!}n^{m-j}\right)(-2)^j = \frac{(-1)^j}{j!(m-j)!}2^mn^m = {m\choose j}(-1)^j\frac{2^mn^m}{m!}$$
and finally, as a polynomial in $n$, $f$ will have leading coefficient $$\sum_{j=0}^m {m\choose j}(-1)^j\frac{2^mn^m}{m!}= \frac{2^mn^m}{m!}\sum_{j=0}^m {m\choose j}(-1)^j = \frac{2^mn^m}{m!}(1 - 1)^m = 0$$
Well, don't that beat all. The leading coefficient from individual terms disappears in the sum.
The 2nd leading term of the individual terms will have degree $m-1$, but its coefficient is more complex. I don't have time to pursue it further tonight, but you can work through the algebra to determine it in a fairly straight-forward manner.
I suspect that the degree $m-1$ terms do not cancel out, which would mean that $f$ grows as $Kn^{m-1}$ as $n \to \infty$ for some constant $K$.