Sum of all possible combinations Guys I just discovered something amazing. Can someone please confirm this? The sum of all possible ways to form a number with $n$ digits, using its digits, without repetition, is equal to $11\ldots1\cdot m(n-1)!$, where $m$ is the sum of the digits of the number, and the amount of $1$'s is equal to $n$. For example, $123$ can be arranged $132, 231, 213, 312, 321$. The sum of these numbers is equal to $1332$. $(111)(6)(2)$. I'll be waiting for my Fields Medal.

Each digit has a "chance" to "occupy" every column. 

Hence the sum is that digit multiplied by $\underbrace{111\cdots 1}_n$.  

Each column will be occupied by every digit, which when summed, gives $m$ for that column.

...Multiply by $m$ to give $\underbrace{111\cdots 1}_n\cdot m$

When a given column occupies a given position, other digits permute
amongst themselves in $(n-1)!$ ways, so the number of times it occupies that column is $(n-1)!$.

...Multiply by $(n-1)!$ to give $\color{red}{\underbrace{111\cdots 1}_n\cdot m (n-1)!}$

