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20 Dec 2009, 15:18
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (03:06) correct
68%
(02:55)
wrong
based on 100
sessions
History
Date
Time
Result
Not Attempted Yet
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
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?
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?
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
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
General Discussion
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.
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.
