The number of ways the skaters can finish the competition is 40,320 ways
The different ways 3 of the skaters finish first, second and third is 56 ways
Given the following
- Number of skaters featured = 8 skaters
If the skaters finish the competition, the number of different ways the skaters finish the competition is expressed as:
8! = 8*7*6*5*4*3*2
8! = 56*30*24
8! = 40,320.
The number of ways the skaters can finish the competition is 40,320 ways
If 3 of the skaters finish first, second and third, the number of ways this can be done is given as:
8C3 = 8!/(8-3)!3!
8C3 = 8!/5!3!
8C3 = 8*7*6*5!/5!3!
8C3 = 56 ways
Hence the different ways 3 of the skaters finish first, second and third is 56 ways
Learn more on combination here: brainly.com/question/11732255
