Hard way or the smart way?

Let us first define some terms,
Sum of given n digits = $S_{n}$
$R_{n} =\underbrace{111\ldots1}_\text{n times}$
Case 1 : Given digits do not include 0.
Suppose we are given n digits, not necessarily unique. Let the number of times a particular digit is repeated be a,b,c,d... and so on.
The total numbers formed using the digits $= \dfrac{n!}{a!b!c!d!\ldots}$
Out of the n given digits, each digit occurs $\dfrac{(n)!}{a!b!c!d!\ldots} \dfrac{1}{n} = \dfrac{(n-1)!}{a!b!c!d!\ldots}$ in the units place, tens place, hundreds place and so on.
The contribution of the one particular digit = $\left(10^{n-1} + 10^{n-2} \ldots 100 + 10 + 1\right) \times \dfrac{(n-1)!}{a!b!c!d!\ldots} \times$ (Value of the digit) = $R_{n}\dfrac{(n-1)!}{a!b!c!d!\ldots}\times$ (Value of the digit )

Therefore sum of all the numbers formed = ( Sum of all digits ) $\times R_{n}\dfrac{(n-1)!}{a!b!c!d!\ldots} = S_{n} \times R_{n}\dfrac{(n-1)!}{a!b!c!d!\ldots}$

Case 2 : Out of n digits, m of them are 0.
Sum of all numbers including those with $0s$ at front = $S_{n} \times R_{n} \dfrac{(n-1)!}{m!a!b!c!d!\ldots} \ldots (1)$

Sum of numbers all numbers formed with $0$ at front = $S_{n} \times R_{n-1} \dfrac{(n-2)!}{(m-1)!a!b!c!d!\ldots} \ldots (2)$
Required sum = $(1) - (2)$
$\therefore$ Required sum = $\dfrac{S_{n}}{a!b!c!d!\ldots} \times \left(\dfrac{R_{n}(n-1)!}{m!} - \dfrac{R_{n-1}(n-2)!}{(m-1)!}\right)$

For the given question
$n = 5, m = 2$ $S_{n} = 9$
$\therefore$ Sum of all numbers formed = $9 \times \left(\dfrac{R_{5}\cdot4!}{2!} - \dfrac{R_{4}\cdot3!}{1!}\right) = 9 \times \left(11111\cdot12 - 1111\cdot6\right) = 9 \times 126666 = 1139994$