In how many ways can you distribute 12 indistinguishable objects into 3 different boxes. where box 1 can have at most 5 objects, box 2 can have at most 6 objects and box 3 can have at most 4 objects
If we were just talking about the question without the inequalities I would use the formula C(n+r-1,n-1).
Since it is not possible I have been thinking if it is possible to answer the question like: $k_{1}+k_{2}+k_{3}=12$ with $k_{1} \leq 5, k_{2} \leq 6, k_{3} \leq 4$
Let $k_1=5-a$, $k_2=6-b$, $k_3=4-c$
Hence the equation converts to $$a+b+c=3$$ Of which we need non negative integral solutions which can be given by $$\binom {5}{2}= 10$$
Note : Luckily none of $a,b,c$ can exceed $3$ if they have to satisfy the equation, thus maintaining $k_1$,$k_2$ and $k_3$ to be non negative
Also there is a generating functions method for such problem which is hinted in one of my answer