Since each student should get atleast 3 prizes, 12 prizes are distributed
(by giving 3 prizes each)
Remaining 4 prizes are to be distributed among the 4 students.
case 1: Any one student gets all the 4 prizes= 4 ways (any of the 4 students)
Case 2: (3,1) or (1,3)
Here any two students are selected and one is given 3 prizes
and other 1 prize = $^4C_2.2=12 \:ways$
Case 3: (2,2)
Here any two students are selected and 2 prizes are given to each.
$^4C_2=6\:ways$
case 4: (1,1,2) or (1,2,1) or (2,1,1)
Here any 3 stdents are selected and given 1 prize to any 2 and 2
prizes to the third one.
This can be done in $^4C_3\times 3=12\:ways$
case 5: (1,1,1,1)
Here each get one prize = 1 way
$\therefore$ The required no. of ways $=4+12+6+12+1=35 \:ways$
