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
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
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