In how many different ways can 8 different books be distributed among 3 students if each receiver gets at least 2 books?
Here's what I did:
First take 6 books from the lot; that can be done in $\binom{8}{6}$ ways.
Then pick two from the 6 books, which can be done in $\binom{6}{2}$ ways. Pick 2 more from the remaining 4, which can be done in $\binom{4}{2}$ ways. The remaining 2 books are taken by the third person.
The remaining two books can go to one person $\binom{3}{1}$ or two different people $\binom{3}{1}\cdot \binom{2}{1}$.
So final answer is $$\binom{8}{6}\binom{6}{2}\binom{4}{2}\binom{2}{2}\left(\binom{3}{1}+\binom{3}{2}\binom{2}{1}\right).$$ Where am I going wrong?
As already explained you are overcounting the correct number: according to your procedure, the same distribution can be obtained in different orders.
Split the enumeration into different cases according to the possible distributions of the books among the three students: say $n_1$ books to the first student, $n_2$ books to the second student and $n_3$ books to the third one.
Since $n_1,n_2,n_3\geq 2$ and $n_1+n_2+n_3=8$, the triple $(n_1,n_2,n_3)$ can be:
1) $(4,2,2)$, $(2,4,2)$ or $(2,2,4)$: we have $3\cdot \binom{8}{4}\cdot \binom{4}{2}$ ways.
2) $(3,3,2)$, $(3,2,3)$ or $(2,3,3)$: we have $3\cdot \binom{8}{3}\cdot \binom{5}{3}$ ways.
Hence the total number is $$3\cdot \binom{8}{4}\cdot \binom{4}{2} +3\cdot \binom{8}{3}\cdot \binom{5}{3}.$$