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How many ways can 12 balls be arranged into 4 different rows

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data

    In how many ways can 12 balls be arranged into 4 different rows with each row having
    at least one ball
    (a) if the balls are identical?
    (b) if there are 6 identical red balls and 6 identical blue balls?

    2. Relevant equations


    3. The attempt at a solution
    a)
    Put 4 balls in each row.
    Remaining 8 balls left.
    Label each row A, B, C, D
    A + B + C + D = 8 balls
    Ans: 11C3 = 165 (Correct)

    b) (Unsure how to do)
     
  2. jcsd
  3. Apr 23, 2016 #2

    jedishrfu

    Staff: Mentor

    You have to look at your first answer and divide out but the number of arrangements.

    As an example, if I have abcd and I want to know all the words I can make the answer is 4!

    But if the d letter is an a then I have to divide my count by 2 a1 b c a2 is the same as a2 b c a1 right?

    So in your what must you do?
     
  4. Apr 23, 2016 #3

    Math_QED

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    Are you sure about this? Divide out? Intuitively, I would say there will be more options because not all balls are identical. I'm not sure though so sorry if I'm mistaken.
     
  5. Apr 24, 2016 #4

    Nathanael

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    May someone please explain to me why the answer to (a) is 11choose3? I agree with it, but I got the answer by way of a nested sum, namely ##\sum\limits_{n=1}^{9}\sum\limits_{m=1}^{n}m## ... I'm curious, what perspective brings out 11choose3?
     
  6. Apr 24, 2016 #5

    Math_QED

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  7. Apr 24, 2016 #6

    haruspex

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    @Math_QED's link is certainly applicable, but here's a more intuitive approach.
    We have n identical objects to place in k distinct buckets. We can place the objects in a line and represent the buckets by drawing boundaries between the objects at k-1 places. We can then think of this arrangement as n+k-1 things, of which some n are the objects. Each possible distribution of the n objects into the k buckets corresponds 1-1 with a decision about which n of the n+k-1 things are objects. Thus the number of distributions is the number of ways of choosing n things from n+k-1 things.
     
  8. Apr 24, 2016 #7

    haruspex

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    Agreed.
     
  9. Apr 25, 2016 #8
    The way I solved it is very similar.

    Assume that you place them in one line and you have 3 brackets (To divide it into 4 groups) So you pretty much have 13 places to place the brackets... However, I cant place them at both ends because it has to have 1 for each group. So 13-2 = 11
    Now choose 3 places from 11
    By
    11c3
     
    Last edited: Apr 25, 2016
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