Too many numbers

To find sum of all numbers with digit sum $S$ & less than $10^{r}$

Consider an r digited number with digit sum S.
Let it's digits be $x_{1} , x_{2} \ldots, x_{r}$
$9 \ge x_{i} \ge 0$
$\therefore x_{1} + x_{2} + x_{3} + \ldots x_{r} = S$
Notice, if we put the first m digits = $0$ we simply get an $r-m$ digit number, so all the numbers are included that is, $1$ digit numbers, $2$ digit numbers, $\ldots$ , r digit numbers.
The number of ways this can be done ( N ) is given by the co-efficient of $x^{S}$ in the expansion of $\underbrace{\left(x^{0} + x^{1} \ldots x^{9}\right)\ldots\left(x^{0} + x^{1} \ldots x^{9}\right)}_\text{r times} = \left(1 + x^{1} + x^{2} \ldots x^{9}\right)^{r} = \left( \dfrac{1-x^{10}}{1-x}\right)^{r}$
$\therefore N = \left(1-x^{10}\right)^{r}\left(1-x\right)^{-r}$
Using the binomial expansion for the first bracket,
$N = \displaystyle \left( \sum_{m=0}^{r}(-1)^{m}\dbinom{r}{m}x^{10m} \right) \times (1-x)^{-r}$
Consider the Taylor series of $\dfrac{1}{1-x}$
$\dfrac{1}{1-x} = \displaystyle \sum_{k=1-r}^{\infty}x^{k+r-1}$

Differentiating $(r-1)$ times,
$\dfrac{(r-1)!}{(1-x)^{r}} = \displaystyle \sum_{k=0}^{\infty}\left((k+r-1)(k+r-2)\ldots(k+1)\right)x^{k}$
$(1-x)^{-r} = \displaystyle \sum_{k=0}^{\infty}\dfrac{\left((k+r-1)(k+r-2)\ldots(k+1)\right)x^{k}}{(r-1)!} = \displaystyle \sum_{k=0}^{\infty}\dbinom{k+r-1}{r-1}x^{k}$
Therefore, co-efficient of $x^{k}$ = $\dbinom{k+r-1}{r-1}$
Therefore co-effcient of $x^{S-10m}$ = $\dbinom{S-10m+(r-1)}{r-1}$
$\therefore$ Co-efficient of $x^{S}$ in $(1-x^{10})^{r}(1-x)^{r}$ = $\displaystyle \sum_{m=0}^{min\left(\left\lfloor\dfrac{S}{10}\right\rfloor, r\right)} (-1)^{m} \dbinom{r}{m} \dbinom{S-10m+(r-1)}{r-1}$

The reason for the upper limit being, $10 \times \left( \left \lfloor\dfrac{S}{10} \right\rfloor +1\right) > S$

$\therefore N = \displaystyle \sum_{\displaystyle m=0}^{min\left(\left\lfloor\dfrac{S}{10}\right\rfloor, r\right)} (-1)^{m} \dbinom{r}{m} \dbinom{S-10m+(r-1)}{r-1}$

For using this in the given question , $S = 24, r = 8$
$\therefore N = \displaystyle \sum_{\displaystyle m=0}^{\left\lfloor\dfrac{24}{10}\right\rfloor} (-1)^{m} \dbinom{8}{m} \dbinom{24-10m+7}{7}$
$\therefore N = \displaystyle \sum_{\displaystyle m=0}^{\displaystyle 2} (-1)^{m} \dbinom{8}{m} \dbinom{31-10m}{7}$

$\therefore N = \dbinom{31}{7} - 8\dbinom{21}{7} + 28\dbinom{11}{7}= 2629575 - 930240 + 9240$
$\therefore N = 1708575$