I am trying to solve the following problem:
Given an integer $n \ge 3$, find an integer $x$ such that:
$${x \choose n} > \frac{x!}{\left(x-\frac{n}{3}\right)!}$$
I don't see any tricks so here's what I came up with.
$$x \ge \sqrt[\frac{2n-3}{3}]{n!}+n-1$$
Here's my thinking:
(1) Assume: $${x \choose n} > \frac{x!}{\left(x - \frac{n}{3}\right)!}$$
(2) Then:
$$x(x-1)(x-2)\dots(x-n+1) > (n!)(x)(x-1)\dots\left(x-\frac{n}{3}+1\right)$$
(3) Since $\dfrac{n}{3} < n$, it follows that:
$$\left(x-\frac{n}{3}\right)\left(x - \frac{n}{3}-1\right)\dots(x-n+1) > n!$$
(4) Since there are $\dfrac{3n-3}{3} - \dfrac{n}{3} = \dfrac{2n-3}{3}$ terms, this is definitely true if:
$$x-n+1 \ge \sqrt[\frac{2n-3}{3}]{n!}$$
Or equivalently:
$$x \ge \sqrt[\frac{2n-3}{3}]{n!}+n - 1$$
Is there a better way to solve this? Is there a more interesting answer to this problem?
$x ≤n$ is the condition for holding relation $\big(^x_n\big)=\frac{n!}{x!(n-x)!}$. If $3|n$ and $\frac{n}{3} ≤x<n$ we may write:
$$\big(^{\frac{n}{3}}_x\big)=\frac{x!}{\frac{n}{3}!(x-\frac{n}{3})!}$$
Or:
$$\frac{x!}{(x-\frac{n}{3})!}=\frac{n}{3}!\big(^{\frac{n}{3}}_x\big)$$
Now we have to show:
$$\big(^x_n\big)>\frac{n}{3}!\big(^{\frac{n}{3}}_x\big)$$
This can be checked numerically, for example:
$n=9$, $x=9/3=3$ gives $\big(^3_9\big)=84$
$\frac{9}{3}!\big(^3_3\big)=6\frac{3!}{3!(0)!}=6$