Find all $4$ digit numbers such that sum of digits is $11$. Find all $4$ digit numbers such that sum of digits is $11$.
\begin{cases} x_1+x_2+x_3+x_4=11\\ 1\leq x_1{\leq 9}\\ 0\leq x_2{\leq 9}\\0\leq x_3{\leq 9}\\ 0\leq x_4{\leq 9}\end{cases}.
Using  stars and bars there is 13 choose 3 ways to do. So we get $286$ but answer is $279$. I saw this same question in the forum but didn't fully understand solution. Where we are overcounting when solving in this way? And mainly what we are overcounting?

In this answer it is described how to solve this by generating functions. The solution is the coefficient of the degree $11$ monomial of the polynomial
$$(x+x^2+\cdots+x^9)(1+x+x^2+x^3+\cdots+x^9)^3$$
This coefficient can be calculated by a CAS like  Mathematica as shown in the link above. It can even calculated by a numerical calculator that can handle arbitrary length integers as describe in the second half of my answer here. But in this answer I learned in one of the comments how to calculate this by hand.
Note that
$$\frac 1 {1-x}=\sum_{n=0}^\infty x^n$$
and by differentiation using induction you get
$$\frac 1 {(1-x)^k}=\sum_{n=0}^\infty{ n+k-1 \choose k-1} x^n$$
So we have
$$(x+x^2+\cdots+x^9)(1+x+x^2+x^3+\cdots+x^9)^3\\
=x\frac{1-x^9}{1-x}\frac{(1-x^{10})^3}{(1-x)^3}\\
=x(1-x^9)(1-x^{10})^3 \sum_{n=0}^\infty{ n+3 \choose 3} x^n\\
=(x-x^{10}-3x^{11}+3x^{20}+\cdots)\frac 1 6 (1\cdot2\cdot3+2\cdot 3 \cdot4\cdot x+\cdots+11\cdot 12\cdot 13\cdot x^{10} \cdots )\\
=\cdots +\cdot \frac 1 6 (-3\cdot 1\cdot 2 \cdot 3-2\cdot 3\cdot 4+11\cdot 12 \cdot 13)x^{11}+\cdots\\
=\cdots+279x^{11}+\cdots$$
The coefficients of the monomials of degree greater than $36$ cancel out.
We have
$$
\frac{\left( {{x}^{40}}-{{x}^{31}}-3 {{x}^{30}}+3 {{x}^{21}}+3 {{x}^{20}}-3 {{x}^{11}}-{{x}^{10}}+x\right) \, \sum_{n=0}^{\infty }{\left. \left( {{n}^{3}}+6 {{n}^{2}}+11 n+6\right)  {{x}^{n}}\right.}}{6}
$$
We abbreviate
$$f(n):={{n}^{3}}+6 {{n}^{2}}+11 n+6$$
and verify that
$$f(n+39)-f(n+30)-3f(n+29)+3f(n+20)+3f(n+19)-3f(n+10)-f(n+9)+f(n)=0$$
and
$$f(n+39)-f(n+30)-3 f(n+29)+3 f(n+20)+3 f(n+19)-3 f(n+10)-f(n+9)$$
for $n=-1,-2,-3$ because $f(n)=0$ for this values
