Combinatorics
posted by Anon .
In how many ways can 8 students be assigned to 2 groups if each group must have at least 3 students?

so the groups could be 3,5 ; 4,4 ; or 5,3
for a total of C(8,3)xC(5,5) + C(8,4)xC(4,4) + C(8,5)xC(3,3)
= 56 + 70 + 56 = 182
I am assuming that the two groups can be distinguished.
If not, then it would simply be 56+70