<s>[INST] 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. 
 [/INST] We wish to find the number of solutions in the nonnegative integers of the equation 
$$a + b + c + d + e = 21 \tag{1}$$
subject to the restrictions $a \leq 3$, $0 < b < 4$, and $c \geq 15$.  Since $b$ is an integer, the restriction $0 < b < 4$ is equivalent to the restriction $1 \leq b \leq 3$.  
We first address the restrictions $b \geq 1$ and $c \geq 15$.  Let $b' = b - 1$ and $c' = c - 15$.  Then $b'$ and $c'$ are nonnegative integers.  Substituting $b' + 1$ for $b$ and $c' + 15$ for $c$ in equation 1 yields
\begin{align*}
a + b' + 1 + c' + 15 + d + e & = 21\\
a + b' + c' + d + e & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with 
$$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$
solutions.  
From these, we must subtract those cases in which $a > 3$ or $b > 3$.  Observe that the restriction $b > 3 \implies b' = b - 1 > 2$.  Notice also that since $4 + 3 = 7 > 5$, these two conditions cannot be violated simultaneously.
Suppose $a > 3$.  Let $a' = a - 4$.  Then $a'$ is a nonnegative integer.  Substituting $a' + 4$ for $a$ in equation 2 yields
\begin{align*}
a' + 4 + b' + c' + d + e & = 5\\
a' + b' + c' + d + e & = 1 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with 
$$\binom{1 + 5 - 1}{5 - 1} = \binom{5}{4}$$
solutions.
Suppose $b' > 2$.  Let $b'' = b' - 3$.  Then $b''$ is a nonnegative integer.  Substituting $b'' + 3$ for $b'$ in equation 2 yields
\begin{align*}
a + b'' + 3 + c' + d + e & = 5\\
a + b'' + c' + d + e & = 2 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with 
$$\binom{2 + 5 - 1}{5 - 1} = \binom{6}{4}$$
solutions.  
Hence, the number of solutions of equation 1 subject to the restrictions $a \leq 3$, $0 < b < 4$, and $c \geq 15$ is 
$$\binom{9}{4} - \binom{5}{4} - \binom{6}{4}$$    
</s>