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.)
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
Jen, check your 1-16, 6:11pm post.
To solve this problem using permutations and combinations, we'll consider two scenarios:
1. Three equal prizes awarded without restrictions:
Since each of the three prizes can be awarded to any of the 20 finalists independently, we can use the concept of combinations.
The number of ways to choose 3 finalists out of 20 without regard to order is denoted as "20 C 3" or "C(20, 3)". This can be calculated using the formula:
C(n, r) = n! / (r!(n-r)!)
Plugging in the values, we have:
C(20, 3) = 20! / (3!(20-3)!) = 20! / (3!17!)
2. First, second, and third place awarded:
In this case, the order of the winners matters, so we need to use permutations.
The number of ways to select the first-place winner is 20. Once the first-place winner is chosen, there are 19 remaining finalists eligible for second place, and once the first and second places are determined, there are 18 finalists eligible for the third place.
Therefore, the total number of ways to award the prizes in the first, second, and third place is:
20 * 19 * 18 = 6840.
So, the answer to your questions are:
1. There are "C(20, 3)" = 1140 ways to award three equal prizes without restrictions.
2. There are 6840 ways to award a first, second, and third place.