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

Please help, Thanks

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    If you assume that the study groups cannot overlap and that every student must be assigned to a group, since you have four groups and at least six people in each group, you will necessarily have seven people in one group and six in each of the rest. Pick who the seven people are in the seven person group. Then left-to-right pick which six people will be in each of the remaining six-person groups. – 2017-02-28

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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.

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    Where did the 4 in - 4• (25C7....so in) , from from? Same with the 18, 12 and additional 6 in the denominater of the 2nd last step? – 2017-02-28
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    @BraydenCarnegie the 4 comes from the choice of which of the four groups is the one to receive the seven people as opposed to six. The 18 comes from the fact that once seven people were selected out of the 25 there are 18 remaining available to select from. Once choosing six of those 18 only 12 remain to select from, etc... As for the representation as a fraction with factorials, that is algebraic manipulation and arithmetic. – 2017-02-28