Number of solutions to equation, range restrictions per variable Find the number of solutions of the equation $x_1+x_2+x_3+x_4=15$ where variables are constrained as follows:
(a) Each $x_i \geq 2.$
(b) $1 \leq x_1 \leq 3$ , $0 \leq x_2$ , $3 \leq x_3 \leq 5$, $2 \leq x_4 \leq 6$
I believe I understand part A. I can fix the values at greater than two by adding two to each term, as in:
$(x_1 + 2) + (x_2 + 2) + (x_3+2) +(x_4+2) = 15$
$x_1+x_2+x_3+x_4=7$
So, $C(7+4-1,7) = {10! \over 7!3!} = 120$
For part B, i really have no idea where to start. Thanks for any insight! 
 A: Assume $a$ as a real number, such that $0<a <1$
The number of integral solutions should be the same as:
The coefficient of $a ^ {15}$ in 
$ \left(a + a^{2} + a^{3} \right)\left(1+a + a^{2} + a^{3} + ... \right)\left(a^{3} + a^{4} + a^{5} \right)\left(a^{2} + a^{3} + a^{4} +a^{5} + a^{6}\right) $
This is same as 
the coefficient of $a ^ {9}$ in 
$ \left(1 + a + a^{2} \right)^{2}\left(1+a + a^{2} + a^{3} + ... \right)\left(1+a+a^{2} + a^{3} + a^{4}\right) $
This is same as 
the coefficient of $a ^ {9}$ in
$  \left(1 -a^{3} \right)^{2}\left(1-a^{5}\right)\left(1-a\right)^{-4} $
Which is same as 
the coefficient of $a ^ {9}$ in
$  \left(1 -2a^{3} + a^6 \right)\left(1-a^{5}\right)\left(1+4a+\frac{4.5}{2!}a^2+\frac{4.5.6}{3!}a^3+\frac{4.5.6.7}{4!}a^4+...\right) $
Which is same as 
the coefficient of $a ^ {9}$ in
$  \left(1 -2a^{3}  + a^6 - a^5+ 2 a^8 \right)\left(1+4a+\frac{4.5}{2!}a^2+\frac{4.5.6}{3!}a^3+\frac{4.5.6.7}{4!}a^4+...\right) $ 
Which is
$\binom{12}{3} - 2\binom{9}{3} + \binom{6}{3} - \binom{7}{3} + 2\binom{4}{3} = 45$
A: By similar reasoning to (a) we can simplify the problem to:
\begin{eqnarray*}
x_1+x_2+x_3+x_4 &=& 9 \qquad\qquad\text{(*)} \\
\text{with } && 0 \leq x_1 \leq 2 \\
&& 0 \leq x_2 \\
&& 0 \leq x_3 \leq 2 \\
&& 0 \leq x_4 \leq 4.
\end{eqnarray*}
Define sets
\begin{eqnarray*}
S\;\, &=& \{\text{all solutions to (*) without upper bounds on the $x_i$}\} \\
S_1 &=& \{\text{all solutions in $S$ where $x_1 \geq 3$}\} \\
S_2 &=& \{\text{all solutions in $S$ where $x_3 \geq 3$}\} \\
S_3 &=& \{\text{all solutions in $S$ where $x_4 \geq 5$}\} \\
\end{eqnarray*}
By the same method used in (a) we calculate:
\begin{eqnarray*}
|S| &=& \binom{9+4-1}{9} = \binom{12}{9} \\
|S_1| = |S_2| &=& \binom{6+4-1}{6} = \binom{9}{6} \\
|S_3| &=& \binom{4+4-1}{4} = \binom{7}{4} \\
|S_1 \cap S_2| &=& \binom{3+4-1}{3} = \binom{6}{3} \\
|S_1 \cap S_3| = |S_2 \cap S_3| &=& \binom{1+4-1}{1} = \binom{4}{1} \\
|S_1 \cap S_2 \cap S_3| &=& 0.
\end{eqnarray*}
Then we require, where set complement is with respect to $S$,
\begin{eqnarray*}
\text{Ans.} &=& |S_1^c \cap S_2^c \cap S_3^c| \\
&=& |S| - |S_1 \cup S_2 \cup S_3| \qquad\qquad\text{by de Morgan's Law} \\
&=& |S| - (|S_1| + |S_2| + |S_3|) + (|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|) - |S_1 \cap S_2 \cap S_3| \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{by the Inclusion-Exclusion Principle} \\
&=& \binom{12}{9} - 2\binom{9}{6} - \binom{7}{4} + \binom{6}{3} + 2\binom{4}{1} - 0 \\
&=& 220 - 168 - 35 + 20 + 8 \\
&=& 45.
\end{eqnarray*}
