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If I have a couple of different sized objects let make it 5 of them and call them 1,2,3,4 and 5. I can place then in boxes differently. There is an unlimited number of boxes. So you could put 1 in a box or 5, does not matter. I can find the number of options of that. If I am correct that should be 2^5-1= 31 ways: object 1 and then object 2, etc, but also 1 and 2, etc...... If I have a list of all these 31 options. My question would be if I were to select the option with the objects 1,3,4 in a box. The other box(es) would have to hold either, object 2 and a box object 5. Or a box with objects 2 and 5.

I hope this is clear enough. What I am looking for is a way to calculate the k (number) of options I have for the n (number) of objects without using a object twice because it is already in a box.

Thank you so much. Can't wait for replies


I believe these would be my options: 1





1 2

1 3

1 4

1 5

2 3

2 4


1 2 3 4 5

Now I want to narrow it down to if I were to pick 1, then I can still have boxes with 2 ... 5 or 2 3

4 5

How many options would there be if you were to pick one and then the next?

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I don't understand why there is an infinite number of boxes. Also, are the boxes different or they are identical and can not be distinguished? Please, clarify. – digital-Ink Oct 19 '12 at 11:26
The reason why I say infinite amount of boxes is because you can then put a object in a box each: box with 1 ,box with 2, box with 3, box with 4, box with 5, or box with 1,2, box with 3,4, box with 5..... I that way. This way I am not limited to the boxes. The boxes are the same, just assume all objects are able to fit into a box. However you can also place just 1 object in there. Hope this clarifies. Please let me know – dave123 Oct 19 '12 at 11:31
If I understand correctly, it means that you want to divide the $n=5$ objects into groups. This is called a partition. There is a formula for the number of ways you can partition a set of $n$ objects into $k$ parts (these are called Stirling's numbers of the second kind). Also the number of ways you can partition a set regardless the number of parts is the sum of Stirling numbers of the 2nd kind for all k from 1 to n and are called Bell's numbers. You can easily search wikipedia for more details. – digital-Ink Oct 19 '12 at 12:03
See my Edit to the question, maybe this helps? – dave123 Oct 19 '12 at 14:09
If I write them out on paper, for boxes 1, 2, 3 I get 5 options. If I write them out on paper, for boxes 1,2,3,4 I get 12 options. Is there an equation for this? – dave123 Oct 19 '12 at 14:23

The number of ways you can distribute $n$ different objects into identical boxes is called Bell's number. You can find detailed information about how you can compute these numbers on Wikipedia. In particular, for $n=5$, there are 52 ways you can distribute the objects (not 31, as you mentioned).

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hi, Sorry if I wasn't clear. But I am assuming that 1,2,3,4,5 is the same as 5,4,3,2,1 or 2,5,4,1,3.... same as 1 2 and 2 1.... etc. Therfore I reach the result of 31 for n=5 – dave123 Oct 21 '12 at 18:44

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