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GRE Math Challenge #17 -A talent contest has 8 contestants
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09 Sep 2014, 03:08
09 Sep 2014, 03:08
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Expert Reply
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Question Stats:
88% (00:37) correct
11% (01:34) wrong based on 18 sessions
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A talent contest has 8 contestants.Judges must award prizes for first,second and third places.If there are no ties,
a) in how many different ways can the three prizes be awarded?
b)how many differnt groups of three people can get prizes?
a) in how many different ways can the three prizes be awarded?
b)how many differnt groups of three people can get prizes?
Show: :: Answer
a) P(9,3) = 8! / (8-3)! = 336
b) C(9,3) = 8! / ( (8-3)! * 3! ) = 56
b) C(9,3) = 8! / ( (8-3)! * 3! ) = 56
Re: GRE Math Challenge #17 -A talent contest has 8 contestants
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20 Jul 2019, 06:39
20 Jul 2019, 06:39
1
a) For this question, we have given that we need to choose the first 3 prizes without the repetition. So the formula is n!/(n-r)!
Where n is the number of things to choose from and we choose r of them, no repetitions, order matters.
So, lets apply it here. 8!/(8-3)!=8!/5!=8*7*6=336
b) How many different groups of 3 ppl can be selected?
The general formula for the number of combinations of r things from n different things is : n!/(n-r)!r!
So, lets apply it here : 8!/5!3!=(8*7*6)/(1*2*3)=56
Where n is the number of things to choose from and we choose r of them, no repetitions, order matters.
So, lets apply it here. 8!/(8-3)!=8!/5!=8*7*6=336
b) How many different groups of 3 ppl can be selected?
The general formula for the number of combinations of r things from n different things is : n!/(n-r)!r!
So, lets apply it here : 8!/5!3!=(8*7*6)/(1*2*3)=56
Re: GRE Math Challenge #17 -A talent contest has 8 contestants
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16 Feb 2020, 18:18
16 Feb 2020, 18:18
Expert Reply
Official question located here https://gre.myprepclub.com/forum/a-talent- ... tml#p45706
Post archived.
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Post archived.
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