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Combinatorics Questions

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data
    How many ways can you select 10 jellybeans from colors Red, Blue, Green so that at most you only have 4 Green jellybeans?

    2. Relevant equations
    ...

    3. The attempt at a solution

    # of ways = # of ways to pick 1 Green + # of ways to pick 2 Green + #of ways to pick 3 Green + # of ways to pick 4 Green.

    1 Green jellybean: After picking out the jellybean, there are then 9 left to choose from.
    * * * * $ * * * * *
    If the '*' are the 9 jellybeans and '$' is the divider to separate the jellybeans so that those on the left of it are Red and those to the right of it are Blue then there are 10!/9! or 10 ways to rearrange it.

    2 Greens: Following the same process above would result in 9!/8! = 9
    3 Greens: 8!/7! = 8
    4 Greens: 7!/6! = 7

    Therefore you would have 10+9+8+7 ways to select 10 jellybeans with at most having 4 Green jellybeans.
     
  2. jcsd
  3. Feb 2, 2015 #2

    Simon Bridge

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    For two greens,
    one comes out first, the second one can come out with the second draw, or the third, or... up to the tenth.
    that's nine ways... but the first could have come out on the 3rd draw, with the second coming out on the 4th or subsequent... thats another 6 ways or something isnt it?
    So that's 15 ways to get 2 green, and I haven't finished counting yet.
     
  4. Feb 2, 2015 #3

    haruspex

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    I would say Extreme112 is interpreting the question correctly, that the order of selection is unimportant.
    You left out one case.
     
  5. Feb 2, 2015 #4
    I think I forgot the 0 case. Thanks for the help guys.
     
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