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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

EDIT:

I believe these would be my options: 1

2

3

4

5

1 2

1 3

1 4

1 5

2 3

2 4

etc

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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    If I write them out on paper, for bo$x$es 1, 2, 3 I get 5 options. If I write them out on paper, for bo$x$es 1,2,3,4 I get 12 options. Is there an equation $f$or this?2012-10-19

1 Answers 1

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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=52012-10-21