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Combinations with Replacement Explanation

  1. Dec 28, 2011 #1
    I'm trying to work out the proof of combinations with replacement, but I cant quite understand the result of it.

    (n+k-1)ℂ(k) is the result, could somebody please explain this to me?
  2. jcsd
  3. Dec 28, 2011 #2


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    Could you state fully and precisely what you want to prove?
  4. Dec 29, 2011 #3
    It is the formula for selection of k items with replacement from a set of n items. I understand the result now, but how do I prove it? Apparently its to be done by induction, but I'm not quite sure of how to approach it.
  5. Dec 29, 2011 #4
    If I understood you correctly your problem is the following:
    Given n items choose k from them. And replacement means you are allowed to choose an item more than once.


    You can interpret the problem in the following way:
    Imagine you have n=3 boxes. Each box stands for a fruit:
    Apple-Box (Box #1)
    Banana-Box (Box #2)
    Orange-Box (Box #3)

    You have k=10 balls and have to throw the balls into the boxes. Suppose you did the following:
    Apple-box: 3 balls
    Banana-box: 2 balls
    Orange-box: 5 balls
    This means you get 3 apples, 2 bananas and 5 oranges.

    So, one combination is:
    In box 1 we have 3 balls
    In box 2 we have 2 balls
    In box 3 we have 5 balls


    We can model this by dividing the 10 balls into 3 blocks:
    o o o | o o | o o o o o
    The blue bars are the dividers.

    Other combinations are:

    All balls in box 1 (only apples):
    o o o o o o o o o o | |

    0 balls in box 1,
    3 balls in box 2,
    7 balls in box 3
    | o o o | o o o o o o o

    1 ball in box 1,
    4 balls in box 2,
    5 balls in box 3
    o | o o o o | o o o o o

    For n blocks (boxes) we need (n-1) dividers, e.g. n=3 requires (n-1)=2 dividers.
    Now, how many "spots" do dividers and balls occupy? We have (n-1) dividers and
    k balls, so in total we have (n-1+k) spots.

    Then, let's analyze how many different positions there are for the balls:
    Since we have (n-1+k) spots and k balls there are (n-1+k)Choose(k) different positions.

    [tex] \binom {n-1+k} {k} [/tex]

    Of course, you could also ask:
    How many different positions are there for the (n-1) dividers?
    Since we have (n-1+k) spots and (n-1) dividers there are (n-1+k)Choose(n-1) different positions. But this is the same as above:

    [tex]\binom{n-1+k}{n-1} = \binom {n-1+k} {k} [/tex]


    Further readings:
    1. Counting selections with replacement
    The derivation is given in the end.

    2. Combinations - order doesn't matter, repetitions allowed
    This is a PF thread.
  6. Dec 29, 2011 #5
    Last edited: Dec 29, 2011
  7. Jan 8, 2012 #6
    Thanks a lot Edgardo! That really helps man!
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