How many combinations are possible?

Question

A box contains n different articles. These articles are to be divided and put into two boxes. It is also possible that all the articles are put in one box. In how many ways this can be done? What will be the case if some of the n articles are identical?

Answer 1

This is like the coin-flipping problem. There are two choices for each item.
Suppose you have 5 items. There are two choices for item #1, then two for item #2, ... so you'll get 2^5 possibilities (if all the items are different). For n items all different, the formula will be...?

If some of the items are the same, then you would think that some of these will be indistinguishable, and so shouldn't be counted as separate arrangements. Suppose you have seven items, but three are the same. Then there would be 2^7 ways if all the items were different. For the four items which are different, there will be 2^4 ways. For the remaining three, there will be either
all 3 in box 1, none in box 2
two in box 1, one in box 2
one in box 1, two in box 2
none in box 1, all three in box 2

this makes four, so the total number of arrangements here would be 4*2^4.

Try some more examples, say with 20 articles, six of which are the same, and you might be able to work out a formula.

Answer 2

Since it doesn't matter which box an item is put into as long as it is put into some box, the answer is n+1 as expressed in other answers above. If however the box identity is important, it becomes a coin flipping problem, as benjamin said. But the word choice suggests box identity is not important.

However, taking a look at bruno's formula.

say you have four identical shirts. Either
Four shirts go into one box.
three shirts go into one and 1 in the other or
two shirts go into both boxes. Thats three combinations... not 5
(4-4+1) *(4+1)

If you have two different items (a,b) and two same items (c) then
either

a and b in one box a and b in separate boxes. < either all c(s) are with a, one c is with a or no c(s) are with a. ( 3 combinations)

that's six combinations. If you think you've found more I believe you're mistaken.

if there's three different things (a,b,c) and two identical items (d) you'd have
(a,b,c) * 3 choices for placement of d
(a,b) * 3 choices for placement of d
(b,c) * 3 choices for placement of d
(a,c) * 3 choices for d

In general it seems that you start adding identical items things become much more complicated.

I must admit, I'm defeated! And a little bit confused. I note that once I've thought about a problem too much I tend to make mistakes really easy, so you may want to take what I've written with a grain of salt the size of Houston.

Thanks for the problem.

Answer 3

There are n+1 ways.

Suppose n is 4. Box A can have 0, 1, 2, 3, or 4 articles. Box B then has n - A articles (i.e. 4, 3, 2, 1 or 0 respectively).

Answer 4

n+1 ways. if some are identical, ie you have 3 a elements, and 3 b elements. then you can put other elements on n+1-6 ways. you can put a elements in 3+1 ways, and b elements in 3+1 ways, so total is (n+1-6)*(3+1)*(3+1)

Answer 5

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