# Combination Problem

1. Sep 6, 2011

### 3.141592654

1. The problem statement, all variables and given/known data

You are given 8 balls, each of a different color. How many distinguishable ways can you:

(1) Divide them (equally or unequally) between 2 urns.
(2) Divide them (equally or unequally) between 2 children (and each child cares about the colors he or she receives).

2. Relevant equations

These are the enumeration formulas we are responsible to know:

Sampling with replacement and order: $n^r$
Sampling without replacement, without order: nCr = $\frac{n!}{r!(n-r)!}$
Sampling without replacement, with order: nPr = $\frac{n!}{(n-r)!}$

3. The attempt at a solution

I initially thought that problem (1) would be without replacement and without order, so that the answer would be a combination with n=8 and r=2, and that problem (2) would be without replacement and with order, so a permutation with n=8 and r=2.

However, that isn't correct. It seems like it might actually be a case where there is replacement. The fact that we are giving the balls to two people, or placing them in two urns, is screwing me up. How can I think about this problem and go about solving it? Is it solvable with just the equations I've listed above? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 6, 2011

### LCKurtz

Think about the 8 balls being lined up in a row.

_ _ _ _ _ _ _ _

Put a 1 in each place if the ball goes in urn A and a 0 if it goes in urn B. How many binary numbers does that give? Then it matters whether the urns are distinguishable.

3. Sep 6, 2011

### 3.141592654

Okay, so if for each ball there would be 2 options for urns, meaning that for the case when the urns are indistinguishable the options available would be:

$2^8 = 256$.

This would be a case where order matters, right? When order doesn't matter I'd have to divide by the number or repetitions, but I'm not understanding how to do that...

4. Sep 6, 2011

### LCKurtz

You mean distinguishable
If you can't tell the urns apart that would just cut it down by half. For example if you put all the balls in urn A and none in urn B, you couldn't distinguish that from its opposite case because you don't know which urn is which.

I'm afraid the wording of the problem is a bit ambiguous regarding the difference between 1 and 2. The balls are all different colors. If in 1 you are supposed to ignore the colors I would think the problem would state that. In any case we have answered 2, assuming the two children aren't identical twins.