Find the number of solutions to this equation using combinatorics 
How many solutions are there to the equation 
  $$x_1 +x_2+x_3+x_4+x_5=21$$ where each $x$ is a non negative integer such that 
   $$0\le x_1 \le3 , 1 \le x_2 \lt4\text{ and }x_3\ge15$$

My attempt : 
So I  first considered the conditions $x_1\ge0 , x_2\ge1$ and $x_3\ge15$
I found out that the total possible combinations with this restriction are $9\choose5$.
But these include do not include the other restrictions on $x_1$ and $x_2$.
So I found out the combinations for the cases of $x_1\ge4$ and $x_2\ge4$ , with an intention of subtracting these from the total combinations of $9\choose 5$.
However I am not ending up with the right answer of $106$.Could you please help me out? I used the formula of $n-r+1\choose r$
 A: The problem should be equivalent to $x_1 +x_2+x_3+x_4+x_5=5$ with $0≤x_1≤3$ and $0≤x_2<3$ and $x_3 \ge 0$.The formula should be $n-r+1\choose {r-1}$ wich gives ${9 \choose 4}=126 $ solutions without the restrictions. Applying the restrictions we need to get rid of the instances of $x_1=4$ (there are 4 of them) and $x_1=5$ (one). Also get rid of $x_2=3$ (by the formula above it is ${2+4-1 \choose 4-1}=10$) and $x_2=4 ($4 solutions$)$ and $x_2=5$ ($1$ solution). Since there is no overlap we can add all these instances and take them out of 126 to give $126-(1+4+10+1+4)=106$
A: The number of solutions of 
$$ x_1+x_2+x_3+x_4+x_5= 5 $$
with the constraints $x_1\in[0,3],x_2\in[0,2]$ and $x_3,x_4,x_5\geq 0$ is given by the coefficient of $z^5$ in the product:
$$ (1+z+z^2+z^3)(1+z+z^2)(1+z+z^2+\ldots)(1+z+z^2+\ldots)(1+z+z^2+\ldots)$$
i.e. by the coefficient of $z^5$ in the Taylor series around $z=0$ of
$$ f(z) = \frac{(1-z^4)(1-z^3)}{(1-z)^5} = (1-z^3-z^4+z^7)\cdot\sum_{n\geq 0}\binom{n+4}{4}z^n $$
i.e. by:
$$ \binom{5+4}{4}-\binom{2+4}{4}-\binom{1+4}{4}=\color{red}{106}.$$
