Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Aug 2016, 03:25

GMAT Club Daily Prep

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Fifteen runners from four different countries are competing

Author Message
TAGS:

Hide Tags

Intern
Joined: 08 Nov 2009
Posts: 47
Followers: 0

Kudos [?]: 41 [1] , given: 0

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

Show Tags

20 Dec 2009, 15:18
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

32% (04:10) correct 68% (02:32) wrong based on 67 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?

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

[Reveal] Spoiler:
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?
[Reveal] Spoiler: OA
Manager
Joined: 25 Aug 2009
Posts: 175
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 12

Kudos [?]: 168 [1] , given: 3

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

Show Tags

20 Dec 2009, 21:57
1
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.
_________________

Rock On

Intern
Joined: 08 Nov 2009
Posts: 47
Followers: 0

Kudos [?]: 41 [0], given: 0

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

Show Tags

21 Dec 2009, 06:23
Atish, thank you for your answer; it seems to make a lot of sense but it does not coincide with the OA.

I'll give everyone some more time to chime in and the I'll post the OA.
Intern
Joined: 22 Dec 2009
Posts: 28
Followers: 0

Kudos [?]: 14 [0], given: 6

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

Show Tags

26 Dec 2009, 05:52
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...
_________________

Deserve before you Desire

Intern
Joined: 25 Mar 2009
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

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

Show Tags

26 Dec 2009, 06:54

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.
Intern
Joined: 09 Sep 2009
Posts: 19
Followers: 0

Kudos [?]: 12 [0], given: 2

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

Show Tags

27 Dec 2009, 10:57
typhoonguywlblwu wrote:

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.
Manager
Joined: 25 Dec 2009
Posts: 66
Followers: 3

Kudos [?]: 33 [0], given: 2

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

Show Tags

28 Dec 2009, 00:41
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!!
Senior Manager
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 333 [0], given: 47

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

Show Tags

05 Jan 2010, 12:18
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!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 500

Kudos [?]: 3055 [2] , given: 360

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

Show Tags

05 Jan 2010, 19:00
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
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)

Marco83, a good question!
+1
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Senior Manager
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 333 [0], given: 47

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

Show Tags

06 Jan 2010, 12:19
Thanks walker... great explanation!
_________________

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

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Senior Manager
Joined: 29 Sep 2009
Posts: 396
GMAT 1: 690 Q47 V38
Followers: 2

Kudos [?]: 31 [0], given: 5

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

Show Tags

21 Nov 2010, 08:05
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?
Senior Manager
Joined: 20 Apr 2010
Posts: 250
WE 1: 4.6 years Exp IT prof
Followers: 8

Kudos [?]: 44 [0], given: 49

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

Show Tags

22 Nov 2010, 12:11
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

Don't Forget to give the KUDOS

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11064
Followers: 511

Kudos [?]: 134 [0], given: 0

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

Show Tags

24 Sep 2014, 00:24
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11064
Followers: 511

Kudos [?]: 134 [0], given: 0

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

Show Tags

04 Aug 2016, 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] 04 Aug 2016, 23:24
Similar topics Replies Last post
Similar
Topics:
1 In the Olympic track represented above 8 runners are going to compete 5 18 Aug 2016, 19:41
4 Four contestants representing four different countries advance to the 8 08 Dec 2014, 06:21
37 There are 24 different four-digit integers than can be 17 04 Nov 2012, 10:39
43 The addition problem above shows four of the 24 different in 12 03 Nov 2010, 00:34
3 On level farmland, two runners leave at the same time from 10 02 Jun 2007, 17:59
Display posts from previous: Sort by