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by harsh.champ » Fri Feb 05, 2010 2:52 am
Two different teams with four students each were training for an international math contest. The students were given an assessment exam and their scores were ranked. What is the probability that the four best scores come from only one of the teams assuming that no two students got the same score?

1. 1/2
2. 1/16
3. 1/35
4. 1/288
5. 1/576
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by thephoenix » Fri Feb 05, 2010 3:56 am
IMO C

P=fav case/total case

tot case=8c4=70
fav case=for the first place there is two choice and rest three places has one choices each
=2C1*1*1*1=2

P=2/70=1/35
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by ajith » Fri Feb 05, 2010 4:01 am
harsh.champ wrote:Two different teams with four students each were training for an international math contest. The students were given an assessment exam and their scores were ranked. What is the probability that the four best scores come from only one of the teams assuming that no two students got the same score?

1. 1/2
2. 1/16
3. 1/35
4. 1/288
5. 1/576
Total num of arrangements = 8!
The arrangements in which all 4 in one team is 1-4 and the second team 5-8 =2*4!*4!

Probability = 2*4!*4!/8! = 48/8*7*6*5 = 1/35
Always borrow money from a pessimist, he doesn't expect to be paid back.
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