Distribute 200 distinct students into 7 groups, where each group has no less than 25 and no more than 29 students.
$$\sum\limits_{i=1}^7x_i = 200 \space \wedge (25\le x_i\le 29)$$
I started with choosing 25 students for each group $${200 \choose 25}{175 \choose 25}{150 \choose 25}{125 \choose 25}{100 \choose 25}{75 \choose 25}{50 \choose 25}*K$$
Then I have to choose the ways the rest of the people can be distributed ($K$):
$$\sum\limits_{i=1}^7y_i = 25 \space \wedge (0 \le y_i \le 4) $$
What is the next step?
Using the multinomial theorem, the number of ways to arrange $27$ people in the first group, $28$ people in the third group, and $29$ people in each of the other groups is $$ \frac{200!}{27!\,29!\,28!\,29!\,29!\,29!\,29!} $$ which is the coefficient of the term $$ 200!\cdot\overset{\text{group $1$}\\}{\frac{x^{27}}{27!}}\cdot\overset{\text{group $2$}\\}{\frac{x^{29}}{29!}}\cdot\overset{\text{group $3$}\\}{\frac{x^{28}}{28!}}\cdot\overset{\text{group $4$}\\}{\frac{x^{29}}{29!}}\cdot\overset{\text{group $5$}\\}{\frac{x^{29}}{29!}}\cdot\overset{\text{group $6$}\\}{\frac{x^{29}}{29!}}\cdot\overset{\text{group $7$}\\}{\frac{x^{29}}{29!}} $$ All similar terms are totaled in $$ 200!\left[x^{200}\right]\left(\frac{x^{25}}{25!}+\frac{x^{26}}{26!}+\cdots+\frac{x^{29}}{29!}\right)^7=3.726481552708924859\times10^{164} $$
Simpler Example
Put $5$ students into two groups where each group has $2$ or $3$ people in it. There are $\binom{5}{2}=10$ ways to put $2$ students in group $1$ and $3$ students in group $2$ and $\binom{5}{3}=10$ ways to put $3$ students in group $1$ and $2$ students in group $2$. That is, $20$ ways to form the groups.
Using the approach in the answer above: $$ \begin{align} 5!\left[x^5\right]\left(\frac{x^2}{2!}+\frac{x^3}{3!}\right)^2 &=120\left[x^5\right]\left(\frac{x^4}4+\frac{x^5}6+\frac{x^6}{36}\right)\\ &=120\cdot\frac16\\[6pt] &=20 \end{align} $$