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.

It's interesting, though not particularly deep. You can just consider the expression for each place of the number (call it $x$). So we first consider the ones' place. If $x$ has decimal representation $x_1x_2\ldots x_n$, then each $x_i$ appears in the ones' place in $(n-1)!$ ways, since if I fix $x_i$ in the ones' place, there are $(n-1)!$ ways to arrange the other $n-1$ digits. This gives $x_1(n-1)!+x_2(n-1)!+\cdots+x_n(n-1)!=(x_1+x_2+\cdots+x_n)(n-1)!$. I'll call $m$ the sum of the digits. I then multiply this by $1$ because it takes the ones' place. I can do the same for the tens' place digit, which gives the same value, but I multiply that by $10$. This goes on for each digit until I have $m(n-1)!+10m(n-1)!+\cdots+10^{n-1}m(n-1)!$ which is exactly what you have.
