Find out the number of solutions to the equation- $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ under certain constraints How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ where each $x_i$ is a non-negative integer, $3 \leq x_1 \leq 10, \ 2 \leq x_2 \leq 7$ and $x_3 \geq 5$
I have been able to do all my counting problems but this one. I can not find an equation for max number per variety.
I know that I will subtract the min for each $x$ from the total, but I can not do $\begin{pmatrix} 15 \\ 6\end{pmatrix}$ cause that would not factor in the max per variety. 
What is the equation I would use for this?
 A: The generating function would be 
$$(x^3+x^4+...+x^{10})(x^2+x^3+...+x^7)(x^5+x^6+...)(1+x+x^2+...)^3$$
Simplify,
$$x^{10}(1-x^8)(1-x^6)({1\over 1-x})^6$$
Now we are finding the coefficient of $x^{25}$ of the above, which is same as  $x^{15}$ in 
$$(1-x^8)(1-x^6)({1\over 1-x})^6=(1-x^8-x^6+x^{14})({1\over 1-x})^6$$
So the coefficient of $x^{15}$ is 
(coefficient of $x^{15}$ in $({1\over 1-x})^6$)$-$(coefficient of $x^{7}$ in $({1\over 1-x})^6$)$-$(coefficient of $x^{9}$ in $({1\over 1-x})^6$)$+$(coefficient of $x^{1}$ in $({1\over 1-x})^6$)
Which is basically
$${15+6-1\choose6-1}-{7+6-1\choose6-1}-{9+6-1\choose6-1}+{1+6-1\choose6-1}$$
A: The answer will be the coefficient of $x^{25}$ from the below expression-
$$(x^3+x^4+ \dots+x^{10})(x^2+x^3+ \dots+x^7)(x^5+x^6+ \dots+x^{20})(x^0+x^1+ \dots+x^{15})(x^0+x^1+ \dots+x^{15})(x^0+x^1+ \dots+x^{15}).$$
A: Let $y_1=x_1-3, \;y_2=x_2-2, \;y_3=x_3-5$, and $y_i=x_i$ for $4\le i\le 6$.
Then $y_1+\cdots+y_6=15$ with $y_1\le7$ and $y_2\le5$;
so if S is the set of solutions in nonnegative integers without restrictions, 
$A_1$ is the set of solutions with $y_1\ge8$, and $A_2$ is the set of solutions with $y_2\ge6$,
$\displaystyle |S|-|A_1|-|A_2|+|A_1\cap A_2|=\binom{20}{5}-\binom{12}{5}-\binom{14}{5}+\binom{6}{5}$
A: Hint: Try to find out the coefficient of $x^{25}$ from the below expression-
$$(x^3+x^4+ \dots+x^{10})(x^2+x^3+ \dots+x^7)(x^5+x^6+ \dots+x^{20})(1+x+\dots+x^{15})^3.$$
That will be your answer.
