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anonymous

  • one year ago

There are K people in a room, each person picks a day of the year to get a free dinner at a fancy restaurant.K is such that there must be at least one group of six people who select the same day. What is the smallest such K if the year is a leap year (366days)? (a) 1829 (b) 1831 (c) 1830 (d) 1832 (e) 1833. Answer is c.please explain how

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  1. anonymous
    • one year ago
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    According to me it is 366*5=1830 but how 1831

  2. mathmate
    • one year ago
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    With 1830, you do not necessarily have a single group of SIX people that choose the same day of the year. With 1831, you will, by the pigeon-hole theorem.

  3. e.mccormick
    • one year ago
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    Ask in math, not computers.

  4. anonymous
    • one year ago
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    Ok can u please explain it by pigeonhole principle....

  5. mathmate
    • one year ago
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    The pigeon hole principle says that if there are n+1 one objects to be placed in n boxes, at least one of the boxes must contain two objects. |dw:1439893780108:dw| Similarly, if you have 366 days, and 1830 people, the "best" you can do to spread it out is to have 366 groups of 5 people, but still will not make a single group of six people. However, if you have 1831 people and 366 days, by the pigeon hole principle, to place the 1831st person, you need to make a group of 6 people, which is the minimum number of people to create at least one group of 6 people.

  6. anonymous
    • one year ago
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    Very helpful thanks a lot

  7. mathmate
    • one year ago
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    You're welcome! :)

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