<s>[INST] How many positive integers less than 1,000,000 have the sum of their digits equal to 19? How many positive integers less than $1,000,000$ have the sum of their digits equal to $19$ ?

I tried to answer it by using stars and bars combinatorics method.
The question says that sum of $6$ digit numbers { considering the number as $abcdef$ } is 19.
Hence, I can write it as $a + b + c + d + e + f = 19$ ,but this assumes that $a,b,c,d,e,f >= 0$ .

But, here each digit lies between 0 and 9. So, how stars and bars method will be modified according to this domain ?
 [/INST] The number is $<$ $1000000$ , 
$\Rightarrow$ it contains 6 digits.
Each of these digits can be one of $0,1,2,3....9$
$\Rightarrow$ problem reduces to no of integral solution to the following equation
$d_1+d_2+d_3+d_4+d_5+d_6 = 19$ where $0\leq d_i \leq 9$
Using generating function :
$$\begin{align*} & \ \ \ \left [ x^{19} \right ]\left ( 1+x+x^2+x^3+....+x^9 \right )^{6} \\ &=\left [ x^{19} \right ]\left [ \frac{1-x^{10}}{1-x} \right ]^{6}\\ &=\left [ x^{19} \right ]\left ( 1-x^{10} \right )^{6}. \sum_{r=0}^{\infty}\binom{5+r}{r}.x^{r} \\ &=\left [ x^{19} \right ]\left [ \sum_{r=0}^{6}.\binom{6}{r}.\left ( -x^{10} \right )^{r} \right ]. \left [ \sum_{r=0}^{\infty}\binom{5+r}{r}.x^{r} \right ] \\ &=\left ( -1 \right )^{0}*\binom{24}{19} + \left ( -1 \right )^{1}*6*\binom{14}{9} \\ &=30492\\ \end{align*}$$
NOTE:
1. $1+x+x^2+x^3+.....x^n = \frac{1-x^{n+1}}{1-x}$
2. $\frac{1}{(1-x)^n} = \sum_{r=0}^{\infty}\binom{n+r-1}{r}.x^r$
3. $\left [ x^{19} \right ]$ means coefficient of $x^{19}$ of the whole expression.
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