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PR1012 Permutations/combinations p17 [#permalink]
 20 Dec 2009, 15:18
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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?
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Re: PR1012 Permutations/combinations p17 [#permalink]
20 Dec 2009, 21:57
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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.
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Re: PR1012 Permutations/combinations p17 [#permalink]
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.
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Re: PR1012 Permutations/combinations p17 [#permalink]
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...
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Re: PR1012 Permutations/combinations p17 [#permalink]
26 Dec 2009, 06:54
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.
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Re: PR1012 Permutations/combinations p17 [#permalink]
27 Dec 2009, 10:57
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.
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Re: PR1012 Permutations/combinations p17 [#permalink]
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!!
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Re: PR1012 Permutations/combinations p17 [#permalink]
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 = 192Thereafter we need to arrange 3 runners from these 4 runners considering the order. Hence its P^4_3The final number should be = 192 * P^4_3 = 4608The answer isn't matching but I guess the method is correct... Can the OA be posted...???
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Re: PR1012 Permutations/combinations p17 [#permalink]
05 Jan 2010, 19:00
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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
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Re: PR1012 Permutations/combinations p17 [#permalink]
06 Jan 2010, 12:19
Thanks walker... great explanation! 
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Re: PR1012 Permutations/combinations p17 [#permalink]
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?
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Re: PR1012 Permutations/combinations p17 [#permalink]
22 Nov 2010, 12:11
Can you Pls give the OA...???
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Re: PR1012 Permutations/combinations p17
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22 Nov 2010, 12:11
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