Question: How many solutions does the following equation have **$ab(c+d+e+f)=33$** (each letter represents a different digit)?

I thought of using "balls($33$) and sticks($4$)" and trying to arrange them but I don't think the answer is $^{36}C\_{33}$.

Answer: 
Assuming $0\leq a,b,c,d,e,f<10$ and you want them to be distinct, then you need $c+d+e+f=11$ and $ab=3$, so $a,b=1,3$ and $c+d+e+f=11$ with $c,d,e,f\in\{0,2,4,5,6,7,8,9\}$. If $c<d<e<f$ then $c+d+e\geq 0+2+4=6$, so $e\leq 5$. That gives you all solutions. 

Then permute to get all solutions.

If they are not distinct, then it is more complicated - you have to consider $a=b=1$ and $c+d+e+f=33$ and you have to consider all ways of writing $c+d+e+f=11$ with any digits.

And if you don't mean digits, well, that's yet another question.
$33=3\cdot 11$, so the three numbers $a, b, c+d+e+f$ must have a product equal to $33$. Consider the cases (like $(1, 1, 33), (1, 3, 11)$ and their permutations) one at a time and use balls and sticks only to determine how many $(c, d, e, f)$'s there are once you have fixed $a$ and $b$.