<s>[INST] 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! 
 [/INST] 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
$\binom{12}{3} - 2\binom{9}{3} + \binom{6}{3} - \binom{7}{3} + 2\binom{4}{3} = 45$
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