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A binomial identity

  1. Apr 25, 2009 #1
    [tex]\sumk=0n\binom{n}{k}2=\binom{2n}{n}[/tex]


    Could someone give me a hint as to how to start this. I'm not sure how to really interpret it.



    [tex](n-k)\binom{n}{k}=n\binom{n-1}{k}[/tex]
    Right Side: Suppose you create a committe from [tex] \binom{n}{k} [/tex], then to pick a leader who isn't in the committee but in the pool of people, we have n-k ways.

    Left Side: Suppose you have n ways to pick a leader for a group. After selecting the leader, you have n-1 people left to pick a committee of size k.
     
    Last edited: Apr 26, 2009
  2. jcsd
  3. Apr 26, 2009 #2

    tiny-tim

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    Hi chaotixmonjuish ! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    The RHS is the number of ways of choosing n people from 2n people.

    Hint: Suppose the 2n people are n men and n women. :wink:
     
  4. Apr 26, 2009 #3
    So would the right hand side be saying that suppose we had n men and n women, there are n ways to form a committee consisitng of both men and women.
     
  5. Apr 26, 2009 #4

    tiny-tim

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    uhhh? :confused:

    the RHS is the same number, no matter how many men (or women) there are.
     
  6. Apr 26, 2009 #5
    Uh oh, ha ha, now i'm confused....I feel like this binomial identiy has some really easy RHS.
     
  7. Apr 26, 2009 #6
    Does it just count the number of ways to form a committee size of n from 2n people?
     
  8. Apr 26, 2009 #7

    tiny-tim

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    Yup! :biggrin:

    Now … pretend the 2n people are n men and n women :wink:
     
  9. Apr 26, 2009 #8
    Okay, so does it still mean n people regardless of gender?
     
  10. Apr 26, 2009 #9

    tiny-tim

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    Yes … the RHS is still the same …

    we wouldn't muck around with that! :rolleyes:
     
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