Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Fifteen runners from four different countries are competing [#permalink]

Show Tags

20 Dec 2009, 14:18

1

This post received KUDOS

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

32% (04:11) correct
68% (02:30) wrong based on 73 sessions

HideShow timer Statistics

Fifteen runners from four different countries are competing in a tournament. Each country holds a qualifying heat to determine who its fastest runner is. These four runners then run a final race for first, second, and third place. If no country has more than one more runner than any other country, how many arrangements of prize winners are there?

My approach is the following, but it doesn't lead to any of the answers:

First, we know that 3 countries have 4 runners, and 1 country has 3. This tells us that there are 4*4*4*3=192 possible final race rosters. Then we can ask ourselves: in how many ways can we arrange 4 runners? 4!=24, hence there are 192*24=4608 possible prize winner arrangements.

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

20 Dec 2009, 20:57

1

This post received KUDOS

If you were to list out all possible scenarios, you would find that each of the 15 runners had an equal probability to come in the 1st, 2nd or 3rd place. We don't need to worry about how they got there (Which country they belong to or did they qualify in the heats). Since the runners are not assigned to any country, they can belong to any country and eventually win the race or come in the 2, or 3rd spot. Hence we simply need to calculate the number of ways in which we can pick 3 people out of 15 with the order being important(Permutation) Hence the answer is 15P3 = 2730.
_________________

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

26 Dec 2009, 04:52

Marco83 wrote:

Fifteen runners from four different countries are competing in a tournament. Each country holds a qualifying heat to determine who its fastest runner is. These four runners then run a final race for first, second, and third place. If no country has more than one more runner than any other country, how many arrangements of prize winners are there?

A) 24 B) 384 C) 455 D) 1248 E) 2730

My approach is the following, but it doesn't lead to any of the answers:

First, we know that 3 countries have 4 runners, and 1 country has 3. This tells us that there are 4*4*4*3=192 possible final race rosters. Then we can ask ourselves: in how many ways can we arrange 4 runners? 4!=24, hence there are 192*24=4608 possible prize winner arrangements.

Where am I making a mistake?

hi Marco83, nice problem... i too m getting an answer that is not provided in the answer stem..

same as your method, but i think we have to multiply the answer by another 4, my answer is 18432.

total 15 members, so 4 from each of 3 countries and 3 from the fourth country... but the country from which 3 participants are competing can be any of the 4 countries...

so selecting 1 winner from each country can be done in, (4*4*4*3)*4 ways= 768 ways

now selecting and arranging (G,S and B) prize winners from the 4 participants can be done in (4C3)(3!) ways=24 ways

hence number of ways of selecting and arranging prize winners= 768*24=18432.

i m not sure about my answer... i hope some senior members of the club can help us...
_________________

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

26 Dec 2009, 05:54

Hi. The answer is 455.

15c3 ...as its a combination problem.No need to worry how the players got the there, as atif correctly mentions..Important consideration is to choose 3 runner from 15 different runners.

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

27 Dec 2009, 09:57

typhoonguywlblwu wrote:

Hi. The answer is 455.

15c3 ...as its a combination problem.No need to worry how the players got the there, as atif correctly mentions..Important consideration is to choose 3 runner from 15 different runners.

I don't think this is the right approach because with this approach it is possible that all the 3 winners are from the same country. However, the question says the last 4 runners are from 4 different countries. Need to use some other approach.

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

27 Dec 2009, 23:41

the answer is 15p3 and not 15c3 because these people could win any of the 3 prizes(arrangement)..if it had said "find out the number of ways in which a person gets a prize" then it would have been a combinaton question

4*4*4*3*4! would give u the distribution..that is the same person get more than 1 prize!!

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

05 Jan 2010, 11:18

atish wrote:

If you were to list out all possible scenarios, you would find that each of the 15 runners had an equal probability to come in the 1st, 2nd or 3rd place. We don't need to worry about how they got there (Which country they belong to or did they qualify in the heats). Since the runners are not assigned to any country, they can belong to any country and eventually win the race or come in the 2, or 3rd spot. Hence we simply need to calculate the number of ways in which we can pick 3 people out of 15 with the order being important(Permutation) Hence the answer is 15P3 = 2730.

atish.. this might not be correct as the question clearly ask us to choose one from each country and then do the final selection of 3 with order in place... hence considering 15 in one go might not be correct...

My approach would be similar to the Marco83's: Only way 4 countries can have the players are - 4,4,4,3 as per the given condition. Ways of selecting final 4 runners \(= C^4_1 * C^4_1 * C^4_1 * C^3_1 = 192\)

Thereafter we need to arrange 3 runners from these 4 runners considering the order. Hence its \(P^4_3\)

The final number should be \(= 192 * P^4_3 = 4608\)

The answer isn't matching but I guess the method is correct...

Can the OA be posted...???
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Marco83 and jeeteshsingh, let's say you have two options:

1) A,B,C,D - runners for final 2) A,B,C,E - runners for final

In both cases you can get A,B,C winners but you counted them twice, are you?

\(P^{15}_3\) is also wrong. Let say we have ABCD runners in the same team. So, ABC combination cannot be real.

My approach: 1. We have runners by countries: 4 4 4 3 2. We choose 3 countries for all 3 "wining" places. 4 possibility: 4 4 4 and 3 times 4 4 3 3. Now we can choose any person from each country and take into account different positions (3!)

So, we get: (4*4*4)*3! + 3*(4*4*3)*3! = 3!*4*4*(4+3*3) = 1248 (D)

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

06 Jan 2010, 11:19

Thanks walker... great explanation!
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

21 Nov 2010, 07:05

I dont see anything wrong with what Marco posted:

First, we know that 3 countries have 4 runners, and 1 country has 3. This tells us that there are 4*4*4*3=192 possible final race rosters. Then we can ask ourselves: in how many ways can we arrange 4 runners? 4!=24, hence there are 192*24=4608 possible prize winner arrangements.

Where am I making a mistake?

As for Walker's explanation - I kind of lost him midway. What are your views guys?

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

22 Nov 2010, 11:11

Can you Pls give the OA...???
_________________

I will give a Fight till the End

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

23 Sep 2014, 23:24

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Fifteen runners from four different countries are competing [#permalink]

Show Tags

04 Aug 2016, 22:24

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...