TITLE: In how many ways we can distribute 7 distinct balls among 3 students such that everyone gets at least 2 balls?
QUESTION [6 upvotes]: So in order to solve this question, I first found all the number of ways in which every student gets at least one ball, i.e. 
$$3^7-3C2 \times (2^7-2) - 3C1=1806$$
and then I subtracted it with number of ways where exactly one ball is given to one or two students, i.e.
$$1806-(3\times7C1(6C2+6C3+6C4))-(3\times7C1\times6C1)=1806-1050-126
=630 ways$$
So can anyone tell is this approach is right? and if you are still not able to understand my answer, please ignore it and solve the question. Thanks in advance

REPLY [6 votes]: There are $\binom{7}{2}$ ways of giving two balls to the first student.
There are $\binom{5}{2}$ ways of giving two balls to the second student.
There are $\binom{3}{2}$ ways of giving two balls to the third student.
Lastly, the last ball can be given to any student, so there are $\binom{3}{1}$ ways to give him.
However, there are repeated cases. For example, the case that we give ball 1&2, 4&5, 6&7 to the first, second, last student respectively and the ball 3 is given to the first student is the same as the case that we give ball 1&3, 4&5, 6&7 to the first, second, last student respectively and the ball 2 is given to the first student. Each case way has appeared $\binom{3}{2}$ times.
Therefore, the answer is $\dfrac{\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{3}{1}}{\binom{3}{2}}=\boxed{630}$