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 $A\_n \to 0$.
Let $[x]$ denote the largest integer not exceeding $x.\;$ For $n\geq 4$ we have $$[n/2]\geq (n-1)/2 \;\text { and } \quad n-[n/2]\geq n/2.$$ Therefore, for $n\geq 4$ we have $$n!\geq \prod\_{j=1+[n/2]}^n j\geq [n/2]^{(n-[n/2])}\geq ((n-1)/2)^{n/2}>1.$$ Hence $n\geq 4\implies n!^{1/n}\geq ((n-1)/2)^{n/2})^{1/n}=\sqrt {(n-1)/2}.$