How many solutions are there to $x_1 + x_2 + ... + x_5 = 21$? How many solutions are there to the equation 

$x_1 + x_2 + x_3 + x_4 + x_5 = 21$ 

where $x_1, i = 1,2,3,4,5$ is a nonnegative integer such that $0 ≤ x_1 ≤ 3, 1 ≤ x_2 < 4$, and $x_3 ≥ 15$?
I have correctly completed the previous parts of the question but am having trouble with this particular restriction.
The total number of solutions to the equation is $C(25,21) = 12650$.
I have tried to do each restriction individually:
For $0 ≤ x_1 ≤ 3$, I solved this by subtracting the number of solutions when $x_1 ≥ 4$ from the total number of solutions. This is $C(25,21) - C(21,17) = 5985$.
For $1 ≤ x_2 < 4$, I have $C(24,20) - C(21,17) = 5985$.
For $x_3 ≥ 15$, I have $C(10,6) = 210$.
However, the restriction is when these are all together. I am not sure how to proceed. The answer is $106$.

The system:
$\begin{cases} x_1+x_2+x_3+x_4+x_5=21\\ 0\leq x_1\leq 3\\ 1\leq x_2<4\\ 15\leq x_3\\ 0\leq x_4\\ 0\leq x_5\end{cases}$
can be rewritten via a change of variables as:
$\begin{cases} y_1+y_2+y_3+y_4+y_5=5\\ 0\leq y_1\color{blue}{\leq 3}\\
0\leq y_2\color{blue}{\leq 2}\\
0\leq y_3\\
0\leq y_4\\
0\leq y_5\end{cases}$
Do you see how?

 Let $x_1=y_1, x_2=y_2+1, x_3=y_3+15$ and $x_4=y_4$ and $x_5=y_5$.  Each of the conditions translate directly as above, including $y_1+y_2+y_3+y_4+y_5=x_1+x_2-1+x_3-15+x_4+x_5=21-1-15=5$

Now, approach via inclusion-exclusion on the first two terms.  If we were to ignore the upper bounds, how many solutions are there?

 $\binom{5+5-1}{5-1}=\binom{9}{4}$

How many of those solutions violate the first upper bound condition?

 that would be when $y_1>3$ i.e. $4\leq y_1$.  Making another change of variable, this would be $z_1+z_2+\dots+z_5=1, 0\leq z_i$ for a total of $\binom{5}{4}=5$

How many of them violated the second upper bound condition?

 Approach similarly

How many violate both upper bound conditions simultaneously?

 That would be with $x_1\geq 4, x_2\geq 4, x_3\geq 15$ implying that $x_1+x_2+x_3+x_4+x_5\geq x_1+x_2+x_3\geq 23$ and couldn't equal $21$, so none violate both.

Letting $|S|$ be the amount which ignore both upper bound conditions, $|A_1|$ be the amount which violate the first upper bound condition, $|A_2|$ be the amount which violate the second upper bound condition, we have the number of solutions which violate none of the upper bound conditions as:
$$|S|-|A_1|-|A_2|+|A_1\cap A_2|$$

 $126 - 5 - 15 + 0 = 106$

