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 $106$.

An approach using generating functions:
In order to count up the number of solutions that sum to $n$, we can look at the coefficient of $x^n$ in the generating function $f(x)$:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{3}x^n\right)\left(\sum_{n=1}^{3}x^n\right)\left(\sum_{n=15}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)$$
The $[x^n]$ operator refers to the coefficient of $x^n$.
Factor out $x$'s so you can shift the indices down to $n=0$ for each sum:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{3}x^n\right)x\left(\sum_{n=0}^{2}x^n\right)x^{15}\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)$$
Use the identities $\displaystyle \sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ and $\displaystyle \sum_{n=0}^{k}x^n = \frac{1-x^{k+1}}{1-x}$ to convert each piece:
$$[x^{n}]f(x) = [x^{n}]\left(\frac{1-x^4}{1-x}\right)x\left(\frac{1-x^3}{1-x}\right)x^{15}\left(\frac{1}{1-x}\right)^3$$
Simplify:
$$[x^{n}]f(x) = [x^{n}]\left(\frac{x^{23}-x^{20}-x^{19}+x^{16}}{(1-x)^5}\right)$$
Factor out $\dfrac{1}{(1-x)^5}$ and use the identity $\displaystyle \frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty}\binom{n+m}{m}x^n$:
$$[x^{n}]f(x) = [x^{n}]\left(x^{23}-x^{20}-x^{19}+x^{16}\right)\sum_{n=0}^{\infty}\binom{n+4}{4}x^n$$
Distribute:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+23}-\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+20}-\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+19}+\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+16}\right)$$
Shift indices:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=23}^{\infty}\binom{n-19}{4}x^{n}-\sum_{n=20}^{\infty}\binom{n-16}{4}x^{n}-\sum_{n=19}^{\infty}\binom{n-15}{4}x^{n}+\sum_{n=16}^{\infty}\binom{n-12}{4}x^{n}\right)$$
If we look at the lower bounds of the indices, the smallest $x^n$ that we could get the coefficient for is $x^{16}$. This is because $16$ is the smallest possible value of $n$ under the $x_i$ constraints, which occurs when $x_1 = x_4 = x_5 = 0$, $x_2=1$, and $x_3=15$. The coefficients for all $n<16$ are simply $0$.
Now we can invoke the coefficient operator:
$$[x^n]f(x) = \binom{n-19}{4}-\binom{n-16}{4}-\binom{n-15}{4}+\binom{n-12}{4}$$
Set $n=21$ to get the number of solutions to the original problem:
$$[x^{21}]f(x) = 106$$
Side note: If you use $n<19$ in the expression above, you'll start to see negative binomial coefficients. For those, you need to use the identity $\binom{-n}{k} = (-1)^k\binom{n+k-1}{k}$
