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A five sportsmen team takes part in swimming competition [#permalink]
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?

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

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.

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.

Re: Counting methods # 27 [#permalink]
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...

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. _________________

Best,

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

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.

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. _________________

Best,

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

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...