<s>[INST] $a+b+c+d+e=79$ with constraints How many non-negative integer solutions are there to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$?
I get that for $a\ge7$ you do $79-7=72$, $\binom{72+5-1}{5-1}=\binom{76}4$. For $b\ge35$ I think it's $\binom{47}4$ and I'm not too sure what it is for $3\le c\le41$ and I also have no clue as to how to do them all at the same time.
 [/INST] So we are looking for non-negative solutions for to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$
I always like, in this kind of problem, to work with generating functions. Everything variable gets a polynomial such that its powers correspond to the restraints, and such that he requested solution would be the coefficient of $x^{79}$ in their product:
The restriction on $a$ translates to the function $$(x^7 + x^8 + x^9 + \ldots) = x^7\left(\frac{1}{1-x}\right)$$ For $b$ we have $$(1+ x + x^2 + \ldots x^{34}) = \frac{1-x^{35}}{1-x}$$ all using standard geometric series. For $c$ we have $$\left(x^3 + x^4 + \ldots + x^{41}\right) = x^3\left(1+x+ \ldots + x^{38}\right) = x^3\left(\frac{1-x^{39}}{1-x}\right)$$ while $d$ and $e$ have no restrictions so we use $$(1+x+x^2 + \ldots) = \frac{1}{1-x}$$
So the answer to your question is the coefficient of $x^{79}$ in:
$$x^7 \frac{1}{1-x} (1-x^{35})\frac{1}{1-x} x^3 (1-x^{39})\frac{1}{1-x}\left(\frac{1}{1-x}\right)^2 $$ which comes down to the coefficient of $x^{69}$ (removing the always present $x^{10}$) in:
$$(1-x^{35})(1-x^{39})(1-x)^{-5} = (1 - x^{35} - x^{39} + x^{74})\sum_{k=0}^\infty {k+4 \choose k} x^k$$ using the generalised binomial formula.
And this coefficient equals $${73 \choose 69} - {38 \choose 34}- {34 \choose 30}$$ 
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