Combination Problem: 8 Balls & 2 Urns/Children

[3.141592654]

Homework Statement

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).

Homework 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)!}

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.

 

[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.

 

[3.141592654]

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...

 

[LCKurtz]

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

You mean distinguishable

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...

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.