In How many ways can $30$ marks be assigned to $8$ problems such that each question has at least $2$ marks?
I solved it by finding the coefficient of $x^{30}$ in $(x^2+x^3\dots)^8$ which turned out to be $\binom{21}7$
But this assumes that the arrangement say $"9,3,3,3,3,3,3,3"$ is same as $"3,9,3,3,3,3,3,3"$
What if the questions are numbered (i.e. the above listed two cases are not the same). How to proceed in that case? I don't think I can just multiply by $8!$ because of various repetition of marks.
If the questions are numbered, consider $$\sum_{i=1}^8 (x_i+2) = 30$$
$$\sum_{i=1}^8 x_i = 14$$
where $x_i \ge 0$.
We can then use a star and bars method to solve the problem.