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. 

Put $b=1+b'$, $c=15+c'$. Then we have to count the nonnegative solutions of
$$a+b'+c'+d+e=5\tag{1}$$
satisfying $a\leq3$, $b'\leq2$. Denote by $n_r$ the number of admissible pairs $(a,b')$ with $a+b'=r$. By stars and bars the total number $N$ of admissible solutions of $(1)$ is then given by
$$N=\sum_{r=0}^5n_r\cdot{7-r\choose 2}=1\cdot21+2\cdot15+3\cdot10+3\cdot6+2\cdot3+1\cdot1=106\ .$$
