The way I interpret the question ...
in the first case you are not worried about the order or actual places, you just want possible groups of 3 people from 20.
That is a combination and it would be C(20,3) or
20!/(3!17!) = 1140
In the second, you are worried about the actual order of first, second and third, so it would be
20x19x18 = 6840
In a mathematics olympiad there are 20 finalists. In how many ways can three equal prizes be awarded, assuming that two prizes cannot be given to one finalist? In how many ways can a first, second and a third place be awarded?
(I understand that Permutations and Combinations have to be used, but I'm not sure how. Please help.)
2 answers
Jen, check your 1-16, 6:11pm post.
3 answers