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** , $\displaystyle a\_n^{1/n}\to 0 \text{ , as } n\to \infty$.
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 $A\_n \to 0$.