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 Post subject: Probability Question #19Posted: Mon Apr 18, 2011 1:50 pm

Joined: Thu Jun 03, 2010 6:58 pm
Posts: 8
Answer explanation says "Since B cannot be ranked, there are 4 teams available to fill the other two spots: 4!/(4-2)! = 12. Multiple 12 x 3 to get 36 permutations that are not allowed"

I got to 12 but I don't know where the 3 comes from? Help plz.

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 Post subject: Re: Probability Question #19Posted: Mon Apr 18, 2011 4:31 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
There are 6*5*4 = 120 possible outcomes if we don't consider the requirement that B has to win a prize if A wins a prize.

We now have to "count" violations of the requirement:

if A is first, there are 4*3 = 12 ways for B to not be 2nd or 3rd
if A is second, there are 4*3 = 12 ways for B to not be 1st or 3rd
if A is third, there are 4*3 = 12 ways for B to not be 2nd or 1st

12 + 12 + 12 = 36

120 - 36 = 84

- Veritas Help

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