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At a recent dog show, there were 5 finalists. One of the finalists was

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Bunuel
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At a recent dog show, there were 5 finalists. One of the finalists was [#permalink] New post   23 Jul 2021, 06:45
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75% (00:38) correct 25% (00:51) wrong based on 84 sessions
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At a recent dog show, there were 5 finalists. One of the finalists was awarded “Best in Show” and another finalist was awarded “Honorable Mention.” In how many ways could the two awards be given out?

A. 5
B. 10
C. 15
D. 20
E. 25
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D
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SD0715
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At a recent dog show, there were 5 finalists. One of the finalists was [#permalink] New post   23 Jul 2021, 08:01
Consider that there are 2 spots : 1st and 2nd. Possible combination of a person to win 1st spot : 5C1 , similarly for 2nd spot , it will be 4C1 ( because same contestant can't win 2nd spot too). Therefore total combinations : 5C1 *4C1 = 5*4 => 20

So the answer is option D, imo

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KetakiUpadhye
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Re: At a recent dog show, there were 5 finalists. One of the finalists was [#permalink] New post   24 Jul 2021, 23:15
Hi SD0715,
Thank you for uploading the solution.... I have just 1 query.... Since in an award order matters could we not have simply calculated the 5P2 in terms of Permutation.

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JannickW
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Re: At a recent dog show, there were 5 finalists. One of the finalists was [#permalink] New post   26 Jul 2021, 00:21
Let X be the winning dog an, Y Honorable mention.

Then I would determine 5 possible ways.

XYYYY
YXYYY
YYXYY
YYYXY
YYYYX

Because there is just one first place and the other ones all do get the same reward we cant just calculate 4*5 right?

I am not sure, please verify.
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Re: At a recent dog show, there were 5 finalists. One of the finalists was [#permalink]
26 Jul 2021, 00:21
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