$a+b+c+d+e=79$ 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.

We may solve this problem by finding the number of solutions with the following constraints 
(1) $a\ge7 $ and $c\ge3$
(2) $a\ge7 $ and $c\ge42$
(3) $a\ge7 $, $b\ge35$ and $c\ge3$
(4) $a\ge7 $, $b\ge35$ and $c\ge42$
The answer to the problem is the answer of (1) - the answer of (2) - the answer of (3) + the answer of (4)
(1) Let $a'=a-7$ and $c'=c-3$. The equation is equivalent to $a'+b+c'+d+e=79-7-3 $, where the variables are nonnegative integers. The number of solutions is $\binom{79-7-3+4} {4 }=\binom {73}{4}$. 
(2) Similar to (1), the number of solutions is $\binom{79-7-42+4} {4 }=\binom {34}{4}$. 
(3) The number of solutions is $\binom{79-7-3-35+4} {4 }=\binom {38}{4}$.
(4) is impossible as $a+b+c>79$.
