## Quiz 3: First Level A Question |
AZobi |
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You and your friend go to the park and meet up with 12 other people at the basketball courts. Two teams of five are randomly selected from the 14 people to play against each other. What is the probability that you and your friend are selected and play on the same team? |

RobinJin: Feb. 7, 2015, 3:22 p.m. |
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I am really not sure if this is right, but the way I thought of the problem as taking the numerator to be the # of ways you could form a team of 5 with you and your friend on the team, so that would be (2 choose 1) to choose the team you and your friend are on and then multiplied by (12 choose 3) to pick the other 3 players. I took the denominator as the total number of ways you could make 2 indistinguishable teams out of 14 people which was (14 choose 5)(9 choose 5) times 1/2 to deorder the groups. Like I said, I'm not too sure if this right so if anyone could help me out that would be pretty cool. |

PMcelhinney: Feb. 7, 2015, 6:05 p.m. |
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I also got the same denominator as RobinJin. However, for my numerator I put (12 choose 3) times (9 choose 5) as the number of ways you and your friend can play on the same team. The first combination chooses only 3 of the 12 people assuming both friends are already on a team together. Then, you choose the second team of 5 out of the 9 people left of the 12. You do not need to account for order in the numerator because there is already an innate order. |

AZobi: Feb. 7, 2015, 7:23 p.m. |
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PMcekhinney, wouldn't the (9 choose 5) in the numerator and the denominator cancel though? |

iraianne: Feb. 8, 2015, 2:17 a.m. |
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Did anyone get (2 choose 2)*(12 choose 3)/ (14 choose 10)? |

weisbart: Feb. 8, 2015, 10:37 a.m. |
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iraianne, what does PMcelhinney get if you simplify the expression? \[\] Also, what happens if you name the teams? If you call one of the teams Team A and the other Team B, what answer do you get? |

Soniakumr: Feb. 8, 2015, 3:55 p.m. |
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I got the same answer in the numerator as you iraianne but in the denominator I wrote (14 choose 5)(9 choose 5) |

dorochi: Feb. 8, 2015, 8:32 p.m. |
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I got something entirely different.... . _ . how I did it was (14 choose 2) for the two friends being chosen, (12 choose 8) for the other 8 people chosen, (10 choose 2) for the friends being chosen for the same team), and (8 choose 3) for the other 3 people on their team. And for the denominator I got (14 choose 5), (9 choose 5) and 1/2 to deorder the teams. |

AZobi: Feb. 8, 2015, 8:48 p.m. |
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I got (12 choose 2)(9 choose 5)(2 choose 2)/ (14 choose 5)(9 choose 5) (1/2!) |

martha2A: Feb. 8, 2015, 9:36 p.m. |
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I got (2 choose 2)(12 choose 3)(9 choose 5) / (14 choose 5) (9 choose 5) (1/2!) I put (12 choose 3) because I am already including me and my friend as being chosen so I subtract us from the 14 and the 5 from what would have originally been (14 choose 5) for the ways 5 could be chosen out of 14. |

JCampos: Feb. 9, 2015, 7:13 a.m. |
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My answer was [(2 C 2)*(12 C 3)]/[(1/2!)*(14 C 5)*(9 C 5) (2 C 2) because both of you will play in the same team (12 C 3) because 12 other players will also play but only 3 of the 12 will be chosen. Notice how the numerator relates to the number of players. 12people + 2 (you&friend) = 14 3 are chosen to play + 2 (you&friend) are chosen to play = 5 then the denominator is the total possible outcomes of how many teams can be formed. You multiply by (1/2!) because the teams are indistinguishable. |

AZobi: Feb. 9, 2015, 11:15 a.m. |
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I'm sorry, I put (12 choose 3) as well. I just typed it incorrectly on here. My answer matches martha2A's. |