<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] 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 $a+b+c>79$.
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