Question: If you count (4 4 3) as one combination, you cannot count (4 3 4) as another.

My approach is $\dfrac{4^3}{3!}$, but obviously this does not work. I don't know why it doesn't work, and I don't know how I should solve the problem.

Answer: 
Another way considering $$A\_n=\left(\frac{1}{n!}\right)^{\frac{1}{n}}$$ Taking logarithms $$\log(A\_n)=-\frac 1n \log(n!)$$ Now, as Shailesh suggested, use [Stirling approximation](https://en.wikipedia.org/wiki/Stirling's_approximation) $$\log(n!)\approx n \log(n)-n+\frac 12 \log(2\pi n)$$ which makes $$\log(A\_n)\approx -\log(n)+1-\frac 1{2n} \log(2\pi n)$$ So, when $n\to \infty$, $\log(A\_n)\to -\infty$ and $A\_n \to 0$.
Using Stirling's Approximation:

$$n!\sim\sqrt{2\pi n}(\frac n e)^n \text{ as } n\to\infty$$

We have:

\begin{align}
\lim\_{n\to\infty}(\frac 1 {n!})^{\frac 1n}&=\lim\_{n\to\infty}[\sqrt{2\pi n}(\frac n e)^n]^{-\frac 1n}\\&=\lim\_{n\to\infty}[(2\pi)^{-\frac 1 {2n}}\cdot n^{-\frac 1{2n}}\cdot\frac en]\\&=\lim\_{n\to\infty} (1\cdot1\cdot0)\\&=0
\end{align}