## Practice Final B2 |
Lilyhui |
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(18 C 7) 7!+(18 C 6)7!+(18C5)*6! |

Lilyhui: March 19, 2015, 6:45 p.m. |
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Here I split the question into three cases: no twin, one twin, and both twins. Another way to approach to problem could be splitting it into two cases: at most one twin and both twin |

rahulchandrupatla: March 19, 2015, 6:47 p.m. |
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that should be right, i did it with the at most one twin and both twin cases which is (19choose7)x7! + (18choose5)6! |

edelapena: March 19, 2015, 6:58 p.m. |
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I got (18C7)*7! + (18C6)*7! + [(18C5)*7!]/2! 0 twins + 1 twin + both twins |

arpanshah: March 19, 2015, 7 p.m. |
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I got the same for the first two cases, but for the third case I got (18C5)7!/2! |

Lilyhui: March 19, 2015, 7:13 p.m. |
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So the additional condition was that if both twins were present they have to stand together which means 7!/2! cannot be used because this includes the situations when they are standing apart. The easiest way to account for their position is treating the twins as one unit which is 6*5!=6! |

arpanshah: March 19, 2015, 7:16 p.m. |
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Actually that does makes more sense, thank you! |

Soniakumr: March 19, 2015, 8:18 p.m. |
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Couldn't you also split it up into 2 cases, where you have at most one twin and then you have both twins. So then it would be (19 choose 7)7! + (18 choose 5) 6! did anyone else do it this way? |

Lilyhui: March 19, 2015, 8:28 p.m. |
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yup! thats also correct! |

rachelobrien: March 19, 2015, 9:40 p.m. |
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why is it multiplied by 6! ? do we not have to divide by 2! to reorder the twins? |

Lilyhui: March 19, 2015, 11:53 p.m. |
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It is 6! because the twins are now one unit (treat as one person since they are required to stand together) Thus since they are seen as one person, no need to deorder them |