Homework - How many non negative solutions? $a+b+c+d+e = 30$
we know that $10\leq e$ and that $4\leq d \leq 7$
How many non negative solutions does this equation have?
 A: That would be the coefficient of $x^{30}$ in 
$$P(x)=(1+x+x^2+...)^3(x^4+x^5+x^6+x^7)(x^{10}+x^{11}+x^{12}+...)$$
Noting we can write $P(x)$ as$\dfrac{x^{14}}{(1-x)^4}(1+x+x^2+x^3) $, we need the sum of coefficients of $x^{13}, x^{14}, x^{15}, x^{16}$ in  $\dfrac1{(1-x)^4}$.  Use the Binomial Theorem for finding these coefficients.  
A: Let $a'=a$, $b\,'=b$, $c'=c$, $d\,'=d-4$, and $e'=e-10$; $30-(4+10)=16$, so you’re trying to count solutions in non-negative integers to 
$$a'+b\,'+c'+d\,'+e'=16\tag{1}$$
with the extra condition that $d\,'\le 3$. Without the extra conditions this is just a stars-and-bars problem; $(1)$ has
$$\binom{16+5-1}{5-1}=\binom{20}4$$
unrestricted solutions in non-negative integers. (Both the formula and a decent derivation of it can be found at the link.) Of course some of these violate the upper bounds on the variables and have to be subtracted from this count. That’s a basic inclusion-exclusion problem; my answer to this question illustrates the method in a similar but slightly more difficult problem.
