Consider
$2l + 3m +4n =11$
I have constraint
$$ 0 \leq l \leq 4$$
$$ 0 \leq m \leq 4$$
and,
$ 0 \leq n \leq 4$
Given this constraint would there be any systematic way to write do all the triple (l,m,n)
Which solve this equation?
Now, after that , would there be a closed from for this summation
$$\sum_{2l + 3m +4n =11} \binom{4}{l} \cdot \binom{7}{m} \cdot \binom{12}{n}$$
My work:
We can immediately figure out true bounds are
$$ 0 \leq m \leq 3 $$ $$ 0 \leq n \leq 2$$ since if they go any higher than that, the sum would overshoot eleven