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: 
**An easy way :** 

Take , $\displaystyle a\_n=\frac{1}{n!}$. Then , $\displaystyle \frac{a\_{n+1}}{a\_n}=\frac{1}{n+1}\to 0 \text{ , as } n\to \infty $.

So by **Cauchy's second limit theorem** , $\displaystyle a\_n^{1/n}\to 0 \text{ , as } n\to \infty$.
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 $n\geq 4\implies n!^{1/n}\geq ((n-1)/2)^{n/2})^{1/n}=\sqrt {(n-1)/2}.$