I was stuck with this particular problem. I tried finding a solution by attempting to find the coefficient of $x^0$ in $(1+x^2 +\dots+x^{10})(1+x^3 +\dots+x^{30})(1+x^{-1} +\dots+x^{-40})$ but for some reason was unsuccessful. would appreciate it if someone could help me out with the approach to tackle this.
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We see that $c=2a+3b$, so the question is essentially asking how many $0\le a\le 5$ and $0\le b\le 10$ there are such that $0\le 2a+3b\le 40$.
For each value of $a$, we must have $b\le (40-2a)/3$ and $b\le 10$, and the number of $b$ which satisfies this is easily counted. One simply needs to add these numbers.
Kenta S
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this is called a lattice. On this site the tag is integer-lattices, and you explore that with https://math.stackexchange.com/questions/tagged/integer-lattices
The set of integers is parametrized as $$ x(1,0,2) + y(0,1,3) $$ That requires proof, but is true. See On a homogeneous Diophantine equation
As we get $$ (x,y,2x+3y) $$ We just need the variables $x,y$ to obey the constraints on your $a,b,$ then check that $2x+3y$ does not exceed $40.$ Not to worry, $$ 2 \cdot 5 + 3 \cdot 10 = 40 $$
Will Jagy
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