## Practice Problem - Lesson 5 |
lindontran |
---|---|

Could someone clarify the example problem at the end of lesson 5 that is labeled "solution"? It begins "Nine people will be selected from a group of 20 for a photograph." I'm not understanding why for Set A, the total number of available people to choose from is 17, when we are choosing at most 1 of the twins and at most 1 of the triplets out of 20. Similarly, for Set B, there are 16 people left to choose from, but we chose both twins and at most 1 triplet out of 20. Why is this? |

lewenchiu: Jan. 22, 2015, 3:37 p.m. |
---|

Set A: "At most 1 of the twins" means that 1 twin definitely cannot be chosen. So, the total number of available people to choose from is reduced to 19. Similarly, "at most 1 of the triplets" means that 2 of the triplets definitely cannot be chosen. So the number of people is then reduced from 19 to 17. Set B: If both twins are already chosen, there are only 18 available people left to choose from. Additionally, "at most 1 triplet" means that the 2 other triplets cannot be chosen, so 18-2=16. |

weisbart: Jan. 22, 2015, 9:42 p.m. |
---|

Nice job lewenchiu! Really nice! |

stephaniehhan: Jan. 23, 2015, 2:19 a.m. |
---|

I have a question relating to this example. If there is a group of 20 people and 9 are chosen, can't it be possible that non of the twins or triplets are chosen? |

weisbart: Feb. 1, 2015, 11:08 a.m. |
---|

Absolutely, but that case is included in the case when at most 1 twin and 1 triplet is chosen... |