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A five sportsmen team takes part in swimming competition

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A five sportsmen team takes part in swimming competition [#permalink] New post 11 Aug 2003, 07:00
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A five sportsmen team takes part in swimming competition with other 20 sportsmen. In how many ways could the members of a team take places in that competition?
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Answer [#permalink] New post 11 Aug 2003, 07:15
(5C1)^2 + (5C2)^2 +(5C3)^2 + (5C4)^2 + 5!
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 [#permalink] New post 11 Aug 2003, 09:28
take places? how many places are there? three as usual? or 25?

There are total 25 people. I assume that all the chances should be considered equal.

25!/5!*20!

Am I right? Am I wrong? Or am I just dreaming?
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Clarification [#permalink] New post 11 Aug 2003, 09:37
stolyar wrote:
take places? how many places are there? three as usual?

There are total 25 people. I assume that all the chances should be considered equal and that there are three prize places.
Thus, each out of 25 people can take 3 places = 3^25

Counting only members of the team 3^5 :roll:


I read the question to mean "In how many ways can the people on the five-person team place in the top five"

This could mean that person 1 places third and his four teammates dont even place.

You are counting every single combination of the placements of all 25 individuals.

What do you think?
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Ignore [#permalink] New post 11 Aug 2003, 09:38
stolyar wrote:
take places? how many places are there? three as usual? or 25?

There are total 25 people. I assume that all the chances should be considered equal.

25!/5!*20!

Am I right? Am I wrong? Or am I just dreaming?


Ignore my comments...your post changed!
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 [#permalink] New post 11 Aug 2003, 09:52
stolyar wrote:
take places? how many places are there? three as usual? or 25?

There are total 25 people. I assume that all the chances should be considered equal.

25!/5!*20!

Am I right? Am I wrong? Or am I just dreaming?


Your answer equals about 53000....does that even make sense for arranging five people? I took this to mean that if _ _ _ _ _ are the five places, the combinations could be:

1 _ _ _ _ or
_ 5 _ _ _ or

_ 3_ 4 1

Where the blanks represent places taken by people NOT on the five person team, but of some of the other 20.

SO...

If only one member places, it could be:

1 _ _ _ _ or
_ 1 _ _ _ or
_ _ 1 _ _ or
_ _ _ 1 _ or
_ _ _ _ 1

Now multiply this by 5 to get all 5 members = 25 combs.

OK, so if two members place, we have more...

1 2 _ _ _
1 _ 2 _ _
1 _ _ 2 _
1 _ _ _ 2
_ 1 2 _ _
_ 1 _ 2 _
_ 1 _ _ 2
_ _ 1 2 _
_ _ 1 _ 2
_ _ _ 1 2

So there are ten here, but we have to multiply this by 5 * 4 to cover all members. So that equals 200.

Now for three we alse have 10 combs (5C3) but we multiply this by 5 * 4 * 3 to get 10 * 60 = 600

For four we get 5C4 = 5 * 5*4*3*2 = 600

for five we get 5! = 120

So lets add these up (my answer has obviously changed)

25 + 200 + 600 + 600 + 120 = 1545

Comments?
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Re: Counting methods # 27 [#permalink] New post 11 Aug 2003, 10:01
Konstantin Lynov wrote:
A five sportsmen team takes part in swimming competition with other 20 sportsmen. In how many ways could the members of a team take places in that competition?


In case "take places" means the first three places, then the answer becomes...

3C1 * 5 = 15 PLUS
3C2 * 5 * 4 = 60 PLUS
3C3 * 5 * 4 * 3 = 60

Answer = 135

See above for a more "complicated" explanation..now I will take a nap...
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answer [#permalink] New post 11 Aug 2003, 23:18
Since places are all unique, it is a promultation problem. The answer is 25A5.
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Re: answer [#permalink] New post 12 Aug 2003, 03:57
Konstantin Lynov wrote:
Since places are all unique, it is a promultation problem. The answer is 25A5.


Another simple way to think of this:

Person1 has 25 possible places
Person2 has 24, etc.

hence, 25x24x23x22x21 or 25P5.
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Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: answer [#permalink] New post 12 Aug 2003, 05:52
AkamaiBrah wrote:
Konstantin Lynov wrote:
Since places are all unique, it is a promultation problem. The answer is 25A5.


Another simple way to think of this:

Person1 has 25 possible places
Person2 has 24, etc.

hence, 25x24x23x22x21 or 25P5.


Akamai,
Great point. Since "to place" in a competition, generally means to come in the top three (gold, silver, bronze) or sometimes five, thats where I was coming from. Your answer is no doubt the best.
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Re: answer [#permalink] New post 12 Aug 2003, 06:12
mciatto wrote:
AkamaiBrah wrote:
Konstantin Lynov wrote:
Since places are all unique, it is a promultation problem. The answer is 25A5.


Another simple way to think of this:

Person1 has 25 possible places
Person2 has 24, etc.

hence, 25x24x23x22x21 or 25P5.


Akamai,
Great point. Since "to place" in a competition, generally means to come in the top three (gold, silver, bronze) or sometimes five, thats where I was coming from. Your answer is no doubt the best.


When making up "GMAT"-type questions, one must be careful not to assume knowledge not generally known by all countries, people, etc. While I will assume everyone knows what dice are, I will always explain a card game, or how chess piece work, or define words (such as "place") that can be ambiguous especially to non-American readers. I try to be very careful as to not make any unwarranted assumptions.

All poster should take note of this and try to edit your problems to be as unambiguous as possible.
_________________

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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: answer   [#permalink] 12 Aug 2003, 06:12
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