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: 
Think about AM-GM, you get  

$(\frac{1}{n!})^{\frac{1}n}\le\frac{2}{n+1}$
We start by writing 

$$\left(\frac1{n!}\right)^{1/n}=e^{-\frac1n \log(n!)}$$

Note that we can easily bound $\log(n!)=\sum\_{k=1}^n\log(k)$ using 

$$n\log(n)-n+1=\int\_1^n \log(x)\,dx\le \log(n!)\le \int\_1^{n+1} \log(x)\,dx=(n+1)\log(n+1)-n$$

Therefore, we have the inequalities

$$-\left(1+\frac1n\right)\log(n+1)+1\le -\frac{\log(n!)}{n}\le -\log(n)+1-\frac1n$$

which obviously show that $\lim\_{n\to \infty}-\frac1n\log(n!)= -\infty$. Finally, we can conclude that 

$$\bbox[5px,border:2px solid #C0A000]{\lim\_{n\to \infty}\left(\frac1{n!}\right)^{1/n}=0}$$

And we are done!