Bounding ${(2d-1)n-1\choose n-1}$ Claim: ${3n-1\choose n-1}\le 6.25^n$.


*

*Why? 

*Can the proof be extended to
obtain a bound on ${(2d-1)n-1\choose
    n-1}$, with the bound being $f(d)^n$
for some function $f$?
(These numbers describe the number of some $d$-dimensional combinatorial objects; claim 1 is the case $d=2$, and is not my claim).

First, lets bound things as easily as possible.  Consider the inequality $$\binom{n}{k}=\frac{(n-k)!}{k!}\leq\frac{n^{k}}{k!}\leq e^{k}\left(\frac{n}{k}\right)^{k}.$$ The $n^k$ comes from the fact that $n$ is bigger then each factor of the product in the numerator.  Also, we know that $k!e^k>k^k$ by looking at the $k^{th}$ term in the Taylor series, as $e^k=1+k+\cdots +\frac{k^k}{k!}+\cdots $. 
Now, lets look at the similar $3n$ and $n$ instead of $3n-1$ and $n-1$.  Then we see that $$\binom{3n}{n}\leq e^{n}\left(3\right)^{n}\leq\left(8.15\right)^{n}$$and then for any $k$ we would have $$\binom{kn}{n}\leq\left(ke\right)^{n}.$$
We could use Stirlings formula, and improve this more.  What is the most that this can be improved?  Apparently, according to Wolfram the best possible is $$\binom{(k+1)n}{n}\leq \left(\frac{(k+1)^{k+1}}{k^k}\right)^n.$$
(Notice that when $k=2$ we have $27/4$ which is $6.25$)
Hope that helps.
