Stars and Bars with bounds 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.

We cannot have $3$ $x_i= > 9$ because the result might be $>= 30$
We can have two $x_i > 9$ because the result may be in the bounds , example $10+10+1+1+1=23$ and $11+10+1+1+0=23$
1 ) let's see the case where there is one or two $x_i > 9$
We can have one $x_i > 9$ because the result may be in the bounds , example $10+9+2+1+1=23$.
The trick would be to remove $10$ and to focalize on the distribution of the remaining $13$ ( which will include a case of twice overflow ) :
Ways to chose the overflow candidates $ = \binom {5}{1} \begin{pmatrix}23 + (5-1) -10 \\ (5-1)\end{pmatrix} = \binom {5}{1} \begin{pmatrix}17 \\ 4\end{pmatrix} = 11900$
2 ) let's see the case where there are two $x_i > 9$ .
Again, the trick is to remove twice 10 and to focalize on the distribution of the remaining $3$. 
Ways to chose the twice overflow candidates $ = \binom {5}{2} \binom {23 + (5-1) -10-10}{5-1} = \binom {5}{2} \binom {7}{4} = 350$
These last combinations had been counted twice and then we must deduced it once.
3 )
Now ignoring the bounds limitation, we have a total of combinations of 
$\begin{pmatrix}23 + (5-1) \\ (5-1)\end{pmatrix} = \begin{pmatrix}27 \\ 4\end{pmatrix} = 17550$
The result must be $17550 - (11900 - 350) =6000$
