The number of ways in which the 8 teams finish found by permutation is 1st, 2nd, 3rd is 336.
A permutation would be a mathematical computation of the number of possible arrangements of a given set, in which the sequence of the arrangements is important.
Now, according to the question,
We have ten teams and want to understand number of times PERMUTATIONS OF 3 are possible. For only a PERMUTATION, we want to consider how several positions we are selecting and how many options are available for each of those positions.
So, given three options are;
(# of choices for first) x (# of choices for second) x (# of choices for third)
If there are originally 8 teams to pick from, then first place has 8 alternatives. When one of those is chosen for first, there are only 7 alternatives for second, and only 6 choices for third... and so on.
8 × 7 × 6 = 336
Therefor, here are 366 possible methods to choose first, second, and third place from the original eight teams.
To know more about permutation, here
https://brainly.com/question/1216161
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