Find the total number of 7-digit Find the total number of 7-digit natural numbers whose digits sum is 20.
Solution: This is not a exactly answer, as well we can use a way to subtracts the cases when exists a digit greater than $9$.
Let $x_1,\cdots, x_7$ the digits.
As well there are ${25\choose 6}$ ways to give us the sums of numbers, but this permits $x_i>9$.
We need to analyze a few cases (or you can abbreviate this if you're awesomely intellectual, I'm not).
$Case~1)$ All the cases (ways) where $x_1>9$ and the others are less than or equal to $9$.
Then you need to write $x_1=9+x_1'$. And why is this a particular case? Because Stars and Bars here admits numbers equal to zero, and in this ways $0$ is permitted.
But see that this case counts ways that is not permitted by our hypothesis, is when$ (x_1,x_i,x_j,\cdots, x_{j+3})=(10,10,0,\cdots,0) $
To calculate the ways it's easy only have that $x_1'+x_2+\cdots +x_7= 11$
Then clearly all the ways are ${16\choose 6} - 6$
$Case~2)$ Some $x_i\neq x_1$ is greater than $9$.
This is only ${15\choose 6} $ but for individual digits, then all the ways are $6 {15\choose 6} $
This give us that the quantity of numbers such that the sum of digits is $20$ are:
$${25\choose 6}-\left({16\choose 6}+6 {15\choose 6}-6\right) $$
Correct

We count the number of  integers $x$ with $1\,000\,000\leq x\leq 9\,999\,999$ which have a digit sum equal to $20$ with the help of generating functions. It is convenient to use the coefficient of operator $[x^{n}]$ to denote the coefficient of $x^n$ in a series.

We obtain
  \begin{align*}
\color{blue}{[x^{20}]}&\color{blue}{\left(1+x+x^2+\cdots+x^9\right)^7-[x^{20}]\left(1+x+x^2+\cdots+x^9\right)^6}\tag{1}\\
&=[x^{20}]\left(\frac{1-x^{10}}{1-x}\right)^7-[x^{20}]\left(\frac{1-x^{10}}{1-x}\right)^6\tag{2}\\
&=[x^{20}]\left(1-7x^{10}+\binom{7}{2}x^{20}\right)\sum_{j=0}^\infty\binom{-7}{j}(-x)^j\\
&\qquad-[x^{20}]\left(1-6x^{10}+\binom{6}{2}x^{20}\right)\sum_{j=0}^\infty\binom{-6}{j}(-x)^j\tag{3}\\
&=\left([x^{20}]-7[x^{10}]+21[x^0]\right)\sum_{j=0}^\infty\binom{j+6}{6}x^j\\
&\qquad-\left([x^{20}]-6[x^{10}]+15[x^0]\right)\sum_{j=0}^\infty\binom{j+5}{5}x^j\tag{4}\\
&=\left(\binom{26}{6}-7\binom{16}{6}+21\binom{6}{6}\right)-\left(\binom{25}{5}-6\binom{15}{5}+15\binom{5}{5}\right)\tag{5}\\
&=\left(230\,230-56\,056+21\right)-\left(53\,130-18\,018+15\right)\\
&=174\,195-35\,127\\
&\,\,\color{blue}{=139\,068}
\end{align*}

Comment:


*

*In (1) we count the strings with length $7$ consisting of digits from $0$ to $9$ with digit-sum $20$. Since numbers do not start with a leading zero, we subtract all strings of length $6$ with digit-sum $20$.

*In (2) we use the finite geometric series formula.

*In (3) we expand the numerator omitting terms which do not contribute to $[x^{20}]$ and we use the binomial series expansion.

*In (4) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.

*In (5) we select the coefficients accordingly.
