Jade S.
asked • 03/24/22What equation should I use to answer this problem? ( nPn = N! or nPr = n! / (n - r)! )
Hello! I am having trouble answering this word problem: Assuming that all 21 contestants are qualified for each of the special awards, in how many ways can the winners for the special awards be chosen?
To clarify, there are three special awards.
Thank you!
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1 Expert Answer
It is not clear from the part of the problem you submitted what is being asked.
If there are 21 contestants each of whom may receive only one award, then you have combinations, NOT permutations.
If each contestant may receive more than one award, you have a much different problem.
Only if the ORDER in which the awards are awarded do you have a problem of permutations.
The number of permutations of 21 things taken 3 at a time is 21P3 = 21*20*19.
Jade S.
Ah, I see. Thank you so much for the clarification! I'm not sure why the question given to our class was structured that way but it is a problem of permutations. So if it is about the order in which the awards are awarded, does that mean I should use the formula nPr = n! / (n - r)! ?
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03/24/22
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Michael M.
So nPn =n! comes directly from the equation nPr = n!/(n-r)!. nPn =n! is just a specific case. Therefore, you always want to use n!/(n-r)!.03/24/22