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