A class has 9 students that are to be put into 3 groups of 3. How many ways possibilities are there?
I'll tell you what it I've tried, but it feels intuitively wrong. There are 9 balls and 3 boxes, and each box takes 3 balls. So the first box has ${9 \choose 3}$ possibilities, and the second box has ${6 \choose 3}$ possibilities because 3 balls are already taken. To find all the possibilities, multiply these and divide by $3!$ because the 3 boxes can be in any order.
$$\frac{{9 \choose 3}{6 \choose 3}}{3!} = \frac{84 \times 20}{6} = 280$$
That seems like way too many for working with such small numbers. Is this the right approach?
EDIT: If this is too easy a question, a general solution to this problem where there are $n$ students and $k$ groups where $k$ divides $n$ would be appreciated.
Here is an alternate solution, although your solution is probably the most straightforward one:
First take one of the 9 people. There are 8 people left, so we have $\binom{8}{2}$ ways to select this person's partners. Now out of the 6 remaining people, take another person at random. We have 5 people left, so there are $\binom{5}{2}$ ways to choose this person's partners.
Thus there are $\binom{8}{2}\binom{5}{2}$ ways to form the 3 groups.