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Probability

by umaa » Sat Oct 03, 2009 6:39 pm
Seven children decide to play "game show," a game in which they select one person to be the host and divide the remaining children into two groups of three. The twins Jeff and Josh dislike each other and won't play on the same team. With that restriction, how many ways are there for the kids to distribute themselves?

(A) 140
(B) 120
(C) 84
(D) 50
(E) 35

OA is D
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by cbenk121 » Sat Oct 03, 2009 7:23 pm
Well here's what I got so far, maybe someone can finish this up:

For the host slot, there are 7 possibilities. For the first team, there are 20 possibilities:

n = 6; k = 3 --> 6! / 3!3! --> 5 x 4 = 20.

For each one of these possibilities, there's only 1 possibility for the second team.

That would be 7 x 20 = 140.

Now how to work in the requirement. I wrote out all the teams that would work, and I got 12 teams. 12 x 7 is 84, so that's my guess of the answer.

However, OA is 50. So I would like to see someone's correct answer to this problem!
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