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Math Help - Combination confusion

  1. #1
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    Combination confusion

    I am having real trouble figuring out when to use these two formulas:

    c(k + t - 1, t-1) and c(k + t - 1, k)

    where k is equal to number of selections
    t is equal to number of elements; number of groups items are in I guess is another way to state that

    For example:

    There are piles of red, green and blue balls; each pile has at least 10 balls in it. How many ways can you select 10 balls?

    I would think it would be c(10 + 3 - 1, 3-1) but the answer the book gives is c(10 + 3 - 1, 10)


    Another one is:

    How many ways can 10 be selected if exactly one red ball must be selected?
    I get c(9 + 2 - 1, 2-1) but again the answer is c(10 +2 -1, 9)

    Please help this confused person!!!!

    You have a bag with twenty balls in it. six red six green and 8 purple


    How many ways can we select 5 balls if balls of each specific color are identical?

    i think this is 20!
    ______
    6!6!8!

    Could you tell me if this is correct? Thanks again folks very much.
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  2. #2
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    \binom {n}{k} = \frac{{n!}}{{\left( {k!} \right)\left( {n - k} \right)!}}=\binom {n}{n-k}
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  3. #3
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    Combination confusion

    Thank you for your quick reply.

    So what you are saying is that c(9 +2-1,9) is the same as c(9+2-1,2-1)????
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  4. #4
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    Quote Originally Posted by Frostking View Post
    Thank you for your quick reply.
    So what you are saying is that c(9 +2-1,9) is the same as c(9+2-1,2-1)????
    Well, of course.
    \binom {9+2-1}{2-1} = \frac{(9+2-1)!}{(2-1)!([9+2-1]-(2-1)])!}
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  5. #5
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    Combination confusion

    PLato, I finally get it. It still does not seem like it should be true but I have tried several using either value k or the number you are choosing - 1 and they are equal. I do not know why it is so hard for me to grasp but I appreciate your patience. Maybe there is someone else out there that was also confused and this will help them before they get as frustrated as I did! Thank s again for the assistance.
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  6. #6
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    Quote Originally Posted by Frostking View Post
    I do not know why it is so hard for me to grasp . Maybe there is someone else out there that was also confused and this will help them before they get as frustrated as I did!
    Did you ask for a combinational proof elsewhere?
    Here is an example. If we have a group of ten from which we choose three then there are seven not chosen. On the other hand, if we choose seven then there are three not chosen. From n if we choose k then there are n-k not chosen and visa-versa.
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