TITLE: Stars and Bars with bounds
QUESTION [8 upvotes]: This question is related to Error solving “stars and bars” type problem

I have what I thought is a fairly simple problem: Count non-negative
  integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$
such that $0 \leq x_1 \leq 9$.

The difference here is on the constraint. It bounds all the $x_i$ under 10 : $\forall i\le5$ ,  $0 \leq x_i \leq 9$.
$8+8+0+0+7=23$ is accepted but not $18+3+0+0+2=23$ or $11+0+0+0+12=23$ . In other words all the $x_i$ must be usual digits.
Note that bad solutions may include one or two bad $x_i$. It is the main difficulty.
What is the count of combinations ?
Edit : this question includes a double bounds and a supplemental difficulty to find the right solution.

REPLY [5 votes]: The result $6000$ provided by @igael is correct. Here is supplement to the answer of @EricStucky. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series. This way we can write e.g.
\begin{align*}
[x^j](1+x)^n=\binom{n}{j}
\end{align*}

We obtain
  \begin{align*}
        [x^{23}]&(1-x^{10})^5\sum_{k\geq 0}\binom{k+4}{4}x^k\\
        &=[x^{23}]\left(1-5x^{10}+10x^{20}\right)\sum_{k\geq 0}\binom{k+4}{4}x^k\tag{1}\\
        &=\left([x^{23}]-5[x^{13}]+10[x^3]\right)\sum_{k\geq 0}\binom{k+4}{4}x^k\tag{2}\\
        &=\binom{27}{4}-5\binom{17}{4}+10\binom{7}{4}\tag{3}\\
        &=17550-5\cdot 2380+10\cdot35\\
        &=6000
\end{align*}

Comment:

In (1) we expand the binomial up to $10x^{20}$ since higher powers do not contribute to $[x^{23}]$.
In (2) we use the linearity of the coefficient of operator and apply the rule $$[x^{p-q}]A(x)=[x^p]x^{q}A(x)$$
In (3) we select the coefficients accordingly.