<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] 
Here is an answer based upon generating functions.
  
  
*
  
*$a\geq 7$ can be encoded as
  \begin{align*}
z^7+z^8+z^9+\cdots=z^7\left(1+z+z^2+\cdots\right)=\frac{z^7}{1-z}\tag{1}
\end{align*}
  
*$b\leq 34$ can be encoded as
  \begin{align*}
1+z+z^2+\cdots+z^{34}=\frac{1-z^{35}}{1-z}\tag{2}
\end{align*}
  
*$3\leq c\leq 41$ can be encoded as
  \begin{align*}
z^3+z^4+\cdots+z^{41}=z^3\left(1+z+z^2+\cdots+z^{38}\right)=\frac{z^3\left(1-z^{39}\right)}{1-z}\tag{3}
\end{align*}
  
*$d,e\geq 0$ can be both encoded as
  \begin{align*}
1+z+z^2+\cdots=\frac{1}{1-z}\tag{4}
\end{align*}

We want to find the number of non-negative integer solutions of
\begin{align*}
a+b+c+d+e=79
\end{align*}
with the constraints given above.

Denoting with $[z^n]$ the coefficient of $z^n$ we are looking for 
  \begin{align*}
[z^{79}]&\frac{z^7}{1-z}\cdot\frac{1-z^{35}}{1-z}\cdot \frac{z^3\left(1-z^{39}\right)}{1-z}\cdot \left(\frac{1}{1-z}\right)^2\tag{5}\\
&=[z^{79}]z^{10}\frac{(1-z^{35})(1-z^{39})}{(1-z)^5}\\
&=[z^{69}]\frac{(1-z^{35})(1-z^{39})}{(1-z)^5}\tag{6}\\
&=[z^{69}]\left(1-z^{35}-z^{39}\right)\sum_{k=0}^\infty\binom{-5}{k}(-z)^k\tag{7}\\
&=\left([z^{69}]-[z^{34}]-[z^{30}]\right)\sum_{k=0}^\infty\binom{k+4}{4}z^k\tag{8}\\
&=\binom{73}{4}-\binom{38}{4}-\binom{34}{4}\tag{9}\\
&=1088430-73815-46376\\
&=968239
\end{align*}
in accordance with the answer of @CYKwong.

Comment:


*

*In (5) we select the coefficient of $[z^{79}]$ of the product of the generating functions (1) to (4) which correspond to the valid ranges specified for $a$ to $e$.

*In (6) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (7) we multiply out the numerator and skip terms with powers greater than $69$ since they do not contribute to $[z^{69}]$. We also apply the binomial series expansion.

*In (8) we use the linearity of the coefficient of operator, apply the same rule as in (6) three times and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.

*In (9) we select the coefficients accordingly.
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