The number of possible outcomes there could be at the end of their game is 6 outcomes
This is a permutation problem since it required arrangement
If Alex, Toby, and Samuel are playing a game together and at the end, they will make a classification with one of them in first place, one of them in Second place and one of them in Third place, this can be done in 3! ways
Since n! = n(n-1)(n-2)!
Hence 3! = 3(3-1)(3-2)
3! = 3 * 2 * 1
3! = 6 ways
Hence the number of possible outcomes there could be at the end of their game is 6 outcomes
Learn more here: https://brainly.com/question/24115376
