In how many ways can 25 students be assigned to 4 distinguishable study groups if at least six people must be in each group?

Question

I'm in Data Management studying combinations, and I don't understand this question.

Please help, Thanks

Answer 1

This problem would be much harder for more students than $25$, but with $25$ students the following technique can be used:

Since each group has at least $6$ students, had they each been of size $6$ that totals $24$ students so the groups must have $7,6,6,6$ students. There are $4$ ways of selecting the group that has $7$ students.

Then there are $\binom{25}{7}$ ways to pick those $7$ students in the special big group.

Then the next group can be chosen in $\binom{18}{6}$ ways and the third group in $\binom{12}{6}$ ways.

So the answer is $$ 4\cdot \binom{25}{7}\cdot\binom{18}{6}\cdot\binom{12}{6}=4\cdot \left(\frac{25!}{18!7!}\right)\cdot \left(\frac{18!}{12!6!}\right)\cdot\left(\frac{12!}{6!6!}\right) = 4\cdot\frac{25!}{7!6!6!6!} $$ which is about 33 trillion.