1. **Problem Statement:** We want to find the number of ways to divide 30 people into 6 groups, each containing exactly 5 people. 2. **Assumptions:** We assume that the groups are unlabeled (i.e., the order of groups does not matter) and that the people are distinct individuals. 3. **Formula Used:** The number of ways to partition $n$ distinct objects into $k$ unlabeled groups each of size $m$ (where $n = k \times m$) is given by: $$\frac{n!}{(m!)^k \times k!}$$ This formula accounts for: - $n!$ ways to arrange all people, - dividing by $(m!)^k$ to account for the indistinguishability of people within each group, - dividing by $k!$ to account for the indistinguishability of the groups themselves. 4. **Applying the formula:** Here, $n=30$, $k=6$, and $m=5$. $$\text{Number of ways} = \frac{30!}{(5!)^6 \times 6!}$$ 5. **Intermediate step showing cancellation:** $$\frac{30!}{\cancel{(5!)^6} \times \cancel{6!}}$$ 6. **Explanation:** This formula ensures that we do not overcount arrangements that differ only by the order of groups or the order of people within groups. **Final answer:** $$\boxed{\frac{30!}{(5!)^6 \times 6!}}$$