Calculate the amount of ways 4 prizes can be awarded to 12 students

[Dell]

calculate the amount of ways 4 prizes can be awarded to 12 students if:
a)each student cannot receive more than one prize
b)each student may receive more than one prize
c)at least one student receives more than one prize

a)
1st prize -> 12 possibilities
2nd prize -> 11 possibilities
3rd prize -> 10 possibilities
4th prize -> 9 possibilities

12*11*10*9 = 11880

b)
1st prize -> 12 possibilities
2nd prize -> 12 possibilities
3rd prize -> 12 possibilities
4th prize -> 12 possibilities

12*12*12*12 = 20736

c)
1st prize -> 12 possibilities
2nd prize -> 11 possibilities
3rd prize -> 10 possibilities
4th prize -> 3 possibilities (the student who won the 1st/2nd/3rd prize)

12*11*10*3=3960

but the correct answer is 8856, which happens to be the subtraction of the answer to b) from the answer to a), why is this the way to solve c) ?? is there not any other way?

 

[Dick]

How about the case where two students get two prizes each? Can you think of some other ways of awarding the prizes your counting doesn't include?

 

[Dell]

now that i look at if after a bit of a break from it, it just seems so obvious, i have b)all the possibilities, and a) the possibilities where nobody wins more than 1 prize, all i need to do is subtract. thanks anyway