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Counting problems: dancing couples & books and children

  1. Apr 12, 2016 #1

    Math_QED

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    1. The problem statement, all variables and given/known data

    1) We choose randomly 10 persons from 10 dance couples, but we need at least one dance couple. In how many ways is this possible?
    2) If we divide n+1 books among n children and every child has to get at least one, in how many ways would this be possible?

    2. Relevant equations

    Combinations, permutations, variations

    3. The attempt at a solution

    1) We want to choose 10 people from 20 people, this is possible in 20 nCr 10 ways.
    We need a couple so we need less than 20 nCr 10 ways. I want to know then:
    20 nCr 10 - the amount of possibilities to have no partner. This is where I got stuck.

    2) We can give n+1 books to n students in (n+1) nPr n ways, but once every pupil has got a book, there is one book remaining. We can give this to any of the students in n ways, but since it doesn't matter in what order this student received the book, we should multiply with 1/2.

    So,[(n+1) nPr n]*n/2 ways

    Can someone verify this?

    Thanks in advance for your quick answers :)
     
  2. jcsd
  3. Apr 12, 2016 #2

    Ray Vickson

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    Is a dancing couple any man paired with any woman, or is it a particular pairing such as Fred and Ethel?
     
  4. Apr 12, 2016 #3

    haruspex

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    If you do not choose both of any pair, how many pairs will you choose one from?
    Yes, but can you simplify that, avoiding the use of nPr. (Also, I don't think this is standard notation. Don't you just mean [(n+1) P n]*n/2?)
     
  5. Apr 13, 2016 #4

    Math_QED

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    1) from 10 pairs one
    2) this is the notation from my graphing calculator, but it's not the standard notation

    Thanks for your reply
     
  6. Apr 13, 2016 #5

    haruspex

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    So how many choices is that?
     
  7. Apr 13, 2016 #6

    Math_QED

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    2^10
     
  8. Apr 13, 2016 #7

    haruspex

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    Indeed.
     
  9. Apr 13, 2016 #8

    Math_QED

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    Thanks for your help :)
     
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