Note: There are a total of $31$ ice cream flavors
I was thinking of approaching this problem by breaking the situation up into cases but now I'm thinking that might be too hectic. For instance if you choose $15$ scoops of any one of the low flavors so ($4$ cases) then you have $10$ scoops left which can come from any combination of the $30$ remaining flavors. So for the leftovers I would do $10$ multichoose $30$ which is $39$ C $30$. Is this the correct approach. I would appreciate feedback on how to solve this more efficiently or critique about what is wrong with my approach. Thank you!
Let $x_k$ be the number of scoops of flavor $k$. If we had no restrictions on the number of scoops of each flavor available, the number of ways we could fill the party bowl would be the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + \ldots + x_{30} + x_{31} = 25 \tag{1}$$ in the nonnegative integers.
Since a particular solution of equation 1 in the nonnegative integers corresponds to the placement of $30$ addition signs in a row of $25$ ones, equation 1 has $$\binom{25 + 30}{30} = \binom{55}{30}$$ solutions since we must choose which $30$ of the $55$ positions required for $25$ ones and $30$ addition signs will be filled with addition signs.
From these, we must subtract those distributions in which there are more than $15$ scoops of chocolate, bubblegum, mint, or strawberry. Notice that it is not possible to simultaneously violate two of these conditions since $2 \cdot 16 = 32 > 25$.
Suppose $x_1$ is the number of scoops of chocolate that are used. If more than $15$ scoops are used, then $x_1' = x_1 - 16$ is a nonnegative integer. Substituting $x_1' + 16$ for $x_1$ in equation 1 yields \begin{align*} x_1' + 16 + x_2 + x_3 + x_4 + \ldots + x_{30} + x_{31} & = 25\\ x_1' + x_2 + x_3 + x_4 + \ldots + x_{30} + x_{31} & = 9 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers with $$\binom{9 + 30}{30} = \binom{39}{30}$$ solutions. These solutions violate the restriction that at most $15$ chocolate scoops are used. By symmetry, this is also the number of solutions of equation 1 that violate the restrictions that at most $15$ bubblegum, mint, or strawberry scoops are used. Hence, there are $$\binom{4}{1}\binom{39}{30}$$ ways to violate the restrictions on the number of available scoops.
Consequently, the number of ways the party bowl can be filled is $$\binom{55}{30} - \binom{4}{1}\binom{39}{30}$$