Answer:
120
Explanation:
You want to know the number of ways the best, second best, and third best could be chosen from a group of 6 essays.
Permutations
The order of choice matters, so we are counting the number of permutations of 3 items chosen from a group of 6.
Best can be any of 6.
Second best can be any of the remaining 5.
Third best can be any of the remaining 4.
The number of ways the top essays could be chosen is 6·5·4 = 120.
Miguel could choose the top essays 120 different ways.
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Additional comment
The function that computes this is often written as nPk, where k items are chosen from a group of n items, and order matters. The function value is ...
nPk = n!/(n-k)!
In this problem, that is ...
6P3 = 6!/(6-3)! = 6·5·4 = 120
Some calculators have statistical functions that will compute this value for you.
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