Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

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).

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.