In how many ways can a teacher divide a group of seven students into two teams each containing at least one student? Can someone please help me with this?
In how many ways can a teacher divide a group of seven students into two teams each containing at least one student? two students? What about when seven is replaced with a positive integer n≥4?
I thought about using combinations.But not sure how to go from there.

well, the basic breakdown goes:
1-6, 2-5, 3-4
For the 1-6, there are 7 choices for the student that is by himself, and the other 6 are dictated by the one by himself, so there are 7 ways to break them up into 2 groups with one student in one and 6 in the other.
For 2-5, you have 7 options for the first person in the group of 2, and then 6 options for the second person, and again -- the other 5 people are the other team.  7*6 means there are 42 ways to break it up into 2 and 5.
For 3-4, you have 7 options, then 6 options, then 5 options for the group of 3, the other 4 people are automatically on the other team.  7*6*5=210 ways to to break it up into 3 and 4.
Hope that helps.
Edit:  What I did above has repeats that I did not consider, so adding a a different, hopefully more useful explanation.
Consider A,B,C,D,E,F,G as the seven people in your group.  If you break the group into to a group of 1 and a group of six, you have seven possibilities (any one of the seven people could be in the group all by themselves.
If you break the group into 2 and 5, as shown above, there are 42 ways to pick 2 people out of a group of 7, but some of those ways are repeats.  Specifically, (A,B) = (B,A).  In fact, every pair of students has a repeat (just picking them in reverse order), so really only half of the 42 methods are different from each other.  So, we only get 21 different ways to break them up into a group of 2 and a group of 5.
If you break them up into a group 3 and 4, there are 210 ways to do this, but again, we get repeats.  Consider picking (A,B,C).  This is the same as picking (A,C,B),  (B,A,C), (B,C,A), (C,A,B), (C,B,A) (this is arranging 3 objects, so we have 3 choices for first place, 2 for the second, and one choice for the last object).  So there are six ways to pick the same group.  In other words, only $\dfrac{1}{6}$ of the groups are unique.  210/6 = 35 different ways to break them up into a group of 3 and a group of 4.
