Given $n$ distinct elements, what is the number of different combinations with unique order? For example, given $1,2,3$, we have 15 combinations:
$1,2,3,12,13,21,23,31,32,123,132,213,231,312,321$.
I have the formula $\sum\limits_{k=1}^{n}\binom{n}{k}*k!$ which can be simplified to $\sum\limits_{k=1}^{n}\frac{n!}{(n-k)!}$
I am wondering whether or not this formula can be further simplified...?
 A: For $n\ge1$, your sum is equal to
$$
\lfloor n!e-1\rfloor
$$
This is because
$$
\begin{align}
\sum_{k=1}^n\frac{n!}{(n-k)!}
&=n!\sum_{k=0}^{n-1}\frac1{k!}\\
&=n!\sum_{k=0}^\infty\frac1{k!}-n!\sum_{k=n}^\infty\frac1{k!}\\
&=n!e-1-\frac1{n+1}-\frac1{(n+1)(n+2)}-\dots
\end{align}
$$
and since $\frac1n=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\dots\gt\frac1{n+1}+\frac1{(n+1)(n+2)}+\frac1{(n+1)(n+2)(n+3)}+\dots$
$$
n!e-1-\frac1n\lt\sum_{k=1}^n\frac{n!}{(n-k)!}\lt n!e-1-\frac1{n+1}
$$

Examples
For $n=2$, this gives $\lfloor5.43656365691809-1\rfloor=4$.
For $n=3$, this gives $\lfloor16.3096909707543-1\rfloor=15$.
For $n=4$, this gives $\lfloor65.2387638830171-1\rfloor=64$.  
A: I don't think the sum you wrote corresponds to the number of combinations in lists like the one you wrote out.
I think you're looking for $$\displaystyle\sum\limits_{k=1}^{n+1} \binom{n+1}{k}$$

This is the sum of the elements in the $(n+1)^\text{th}$ row of Pascal's Triangle, minus one. This is equal to $$2^{n+1}-1$$. 
For example, when $n=3$, this gives $2^4-1 = 15$.  

EDIT: This answer is incorrect (see comments).
