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Unique ways to keep N balls into K Boxes?

This question may be sound stupid, but we really cant figure it out. We have 3 rocks and 6 boxes. All the rocks have to be in the boxes. The rocks can be all in one box or spread out.

How many unique combinations can we possible have?

I made a spreadsheet of this https://docs.google.com/spreadsheet/ccc?key=0AjBAKweB5syRdDFVeE5qNVJqT3F5RE9heERvYVBWdnc&hl=en_US

i came up with 53. I am missing some? please let me know.

The question to this problem is whats the equation. How do we come up with this number without the spreadsheet so i can apply it to a another set of questions.

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marked as duplicate by Arturo Magidin, sdcvvc, Matt N., Jyrki Lahtonen, Quixotic Sep 4 '12 at 21:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Can you tell the boxes apart? Can you tell the rocks apart? So, for example, all rocks in box 1 is the same (or different) arrangement as all rocks in box 2? –  GEdgar Sep 14 '11 at 21:05
    
@GEdgar rocks dont matter they have no uniqueness, but the boxes do. box 1, box 2, box 3 .... so in your case it would be different since each box has a value. –  M. of CA Sep 14 '11 at 21:07
2  
This is an instance of the stars and bars problem. See here or here. –  Arturo Magidin Sep 14 '11 at 21:20
    
@Arturo i dont think stars and bars that repetitiveness in account. Correct me if i am wrong. –  M. of CA Sep 14 '11 at 21:30
    
@M. of CA: Sure it does; you just have to interpret it correctly. Put the three rocks down, * * *; now place five vertical lines to indicate where to go to the next box. E.g., ||**|*|| means first two boxes are empty, third one has two rocks, fourth one has one rock, last two are empty. It's a "combinations with repetitions" problem, and those problems are always equivalent to stars and bars. See also combinations with repetitions –  Arturo Magidin Sep 14 '11 at 21:33

1 Answer 1

up vote 0 down vote accepted

It all depends on what you count as a different pattern.

If each rock and each box count as different you can put each rock in one of six boxes so you get $6^3$. The number of ways of having each rock in a different box is $6\times 5 \times 4$, so the probability is $\frac{5}{9}$.

If the rocks all look the same and the boxes look the same there are $3$ ways: three boxes with one rock each and three empty; a box with two, a box with one and four with none; or a box with three and five with none. The probability of having each rock in a different box becomes $\frac{1}{3}$ if each pattern is equally likely.

Or perhaps the boxes look the same and the rocks different, or the boxes are different but the rocks look the same, and you have two more possible answers. You seem to want the last of these, and it is not difficult, so why not show us what you think the answer might be.

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I think the answer is 53. the boxes are differnt the rocks are not. –  M. of CA Sep 14 '11 at 21:12
    
I don't think it is 53, though that is within 10% of my answer. How did you come to 53? –  Henry Sep 14 '11 at 21:21
    
please check this link docs.google.com/spreadsheet/… –  M. of CA Sep 14 '11 at 21:23
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@M. of CA: Your count is incomplete; where is the distribution that has 1 rock in box 4, and 2 in box 5? 1 in box 4, 2 in box 6? 1 in box 5, 2 in box 6? –  Arturo Magidin Sep 14 '11 at 21:29
    
@Arturo Thank you... I have fixed this. i missed of skipped it. –  M. of CA Sep 14 '11 at 21:34

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