How many solutions does the equation $a + b + c + d + e = 21$ have in the nonnegative integers if $a \leq 3$, $0 < b < 4$, and $c \geq 15$? How many solutions are there to the equation in the nonnegative integers:
$$a+b+c+d+e = 21$$
Conditions:
a) $ a \le 10$
I understood the solution which is total number of possibilities - the number of possibilities with $a \gt 10$ which is $\binom{5+21-1}{21} - \binom{5+10-1}{10}$.
b) $a \le 3$, $0 \lt b \lt 4$, $c \ge 15$.
How do I solve this part? Thanks in advance. 

If you don't mind, I'll solve this with generating functions. We represent the different choices for each letter as a power series, where the coefficient represents the allowed numbers. In your case, $a$ will be represented by $(1+x+x^2+x^3)$ since it can be $0,1,2,3$, similarly $b$ will be $(x+x^2+x^3)$ , $c$ will be $\sum_{k=15}^\infty x^k = x^{15}\sum_{k=0}^\infty x^k$, and $d$ and $e$ will be $\sum_{k=0}^\infty x^k$ since they can be anything. Now, when we multiply these power series together, for every way that we can choose a term from the letters and make the power $21$, we will add $1$ to the coefficient of $x^{21}$ in the product power series. 
To convince yourself that this works, you could look at the generating function $(1+x)^n = \sum_{k=0}^n\binom{n}kx^k$. This is like having $n$ letters that can be $0$ or $1$, and the coefficient of $x^k$ in the product power series is the number of ways which you can write $k$ as sum of $0$s and $1$s, which is just the number of ways which you can choose which $k$ of the $n$ variables to take on the value $1$. 
Now, our product series is
\begin{align*}
(1+x+x^2+x^3)x(1+x+x^2)x^{15}\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=0}^\infty x^k\right)^2 &= x^{16}\frac{1-x^4}{1-x}\frac{1-x^3}{1-x}\frac1{(1-x)^3} \\
&= \frac{x^{16}(1-x^4)(1-x^3)}{(1-x)^5}.
\end{align*}
So, we want the coefficient of $x^{21}$ of the power series expansion of the above, which is just the coefficient of $x^5$ in the expansion of
$$
\frac{(1-x^4)(1-x^3)}{(1-x)^5} = (1-x^3-x^4+x^7)\sum_{k=0}^\infty\binom{k+4}4 x^k.
$$
The relevant terms come from $1\cdot\binom{5+4}4 x^5$, $-x^3\cdot\binom{2+4}4 x^2$, $-x^4\cdot\binom{1+4}4 x$, so your answer is
$$
\boxed{\binom{5+4}4-\binom{2+4}4-\binom{1+4}4.}
$$
Please let me know if I have made any mistakes. 
