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Judges will select 5 finalists from the 7 contestants

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Judges will select 5 finalists from the 7 contestants [#permalink] New post 17 Jan 2005, 13:39
Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible?

a) 10
b) 21
c) 210
d) 420
e) 1,260
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 [#permalink] New post 17 Jan 2005, 14:10
E. 1260 ways.

First select 5 ppl from the gp . 7C5 and then rank them in 5 x 4 x3 ways


Total 7C5 * 5x4x3 = 21 x 60 = 1260 ways
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 [#permalink] New post 17 Jan 2005, 18:50
gayathri wrote:
OA is C


what's the explanation Gayathri ?
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 [#permalink] New post 17 Jan 2005, 18:53
Gayathri can you please post the explanation
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 [#permalink] New post 17 Jan 2005, 19:29
Sorry, dont have one.
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 [#permalink] New post 18 Jan 2005, 02:42
7C5.5C3 = 6.7.4.5/4=5.6.7 = 210

Then 210 is the result of the question "how many groups of winners do exist", not in how many arrangements can we rank this 5 guys

Hope it helps
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 [#permalink] New post 18 Jan 2005, 05:24
Thanks for the explanation Twixt.

I also figured another approach to this. The whole logic of selecting 5 out of 7 is just to throw us in a loop. In the end, there will be three candidates from the 7 to get the prizes.

However since order matters, its a permuation issue.
i.e: 7 possible choices for 1st place, 6 for second and 5 for third.

So total possible arrangement = 7*6*5=210
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 [#permalink] New post 18 Jan 2005, 06:08
gayathri wrote:
Thanks for the explanation Twixt.

I also figured another approach to this. The whole logic of selecting 5 out of 7 is just to throw us in a loop. In the end, there will be three candidates from the 7 to get the prizes.

However since order matters, its a permuation issue.
i.e: 7 possible choices for 1st place, 6 for second and 5 for third.

So total possible arrangement = 7*6*5=210

thats an easier and much clearer route to take.. Gayathri...thanks..
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 [#permalink] New post 18 Jan 2005, 10:08
The choice is between 5 people and not 7 people hence 1260 is the right answer as there will be 3 canidates from 5 people
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 [#permalink] New post 20 Jan 2005, 08:40
i fully agree with sunju...the answer must be 1260.
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 [#permalink] New post 23 Jan 2005, 17:44
Well i can explain why the answer is 210.
The principle of choice is being tested here and i think here it goes:

First out of 5 finalist out of 7 contestants are selected, meaning order does not matter it's just 5 people from 7 people thus we use 7C5 and we get 21

Then out of these 5, 3 contestants are ranked such that: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place (this means order matters because it asks for different "arrangement" of prize winners) thus we use 5P3 and we get 10.

21 * 10 = 210
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 [#permalink] New post 23 Jan 2005, 23:42
It's a trick question. Out of the seven people only three are going to get prizes, the other two doesn't matter. So 7P3=210.
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 [#permalink] New post 24 Jan 2005, 05:26
Folaa,

I thought 5P3 = 20 not 10
5C3 = 10
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 [#permalink] New post 24 Jan 2005, 10:21
5P3=5*4*3=60? Or am I totally confused?
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 [#permalink] New post 29 Apr 2005, 08:52
Folaa3 wrote:
Well i can explain why the answer is 210.
The principle of choice is being tested here and i think here it goes:

First out of 5 finalist out of 7 contestants are selected, meaning order does not matter it's just 5 people from 7 people thus we use 7C5 and we get 21

Then out of these 5, 3 contestants are ranked such that: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place (this means order matters because it asks for different "arrangement" of prize winners) thus we use 5P3 and we get 10.

21 * 10 = 210


Folaa3, what do u think if within 3 contestants A,B,C, the order can change, first A, second B, third C or first B, second C, third A. So # of ways to arrange 3 best winners are 3!

The total ways = 7C5*5C3*3! = 1260
  [#permalink] 29 Apr 2005, 08:52
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