Distributing 5 Objects to 3 Boxes: C(5,3)

AI Thread Summary
The discussion centers on the problem of distributing five distinguishable objects into three indistinguishable boxes. Participants debate whether to use the binomial coefficient C(5,3) or to treat the boxes as distinguishable and then adjust for indistinguishability by dividing by 3!. It is clarified that multiple objects can fit into a box, and the correct approach involves considering the number of ways to partition the objects into subsets. The final consensus indicates that the total number of distributions is 46, derived from specific groupings rather than using C(5,3). The conversation concludes with a suggestion that the formula for calculating these distributions is improving.
gimpy
Messages
28
Reaction score
0
I think i got this answer.

How many ways are there to distribute five distinguishable objects into three indistinguishable boxes?

Wouldn't the answer just be C(5,3) because the boxes are indistinguishable? Or do i treat this question the same as if the boxes were distinguishable?
 
Mathematics news on Phys.org
The binomial coefficient C_3^5 is right. You could do it as if the boxes were distinguishable and then divide by the number of sequences of them: 3! = 6. and that's what you would get.
 
Multiple objects can fit in a box, can't they?
 
Unless they are very, very big.
 
Originally posted by Hurkyl
Multiple objects can fit in a box, can't they?

Yes, multiple objects can fit into a box.

I think my answer is correct, isn't it?
 
I don't think it is. I think it's quite a way out. The answer is the same as the number of ways to partition 5 into three (possibly empty?) subsets, which is

\sum\binom{n}{p}\binom{n}{q}\binom{n}{r}

for all p+q+r = 5 (and possibly with the requirement that p,q,r are strictyl positive.

To show that your answer is wrong, I think your reasoning that the number of ways to put 5 distinct objects into 1 box is 5 choose 1, ie 5, when obviously it is 1. Also there are many ways of putting 5 balls into 100 boxes, and 5 choose 100 is define to be 0.
 
There are 5 possible groupings:
0 0 5 - 1
0 1 4 - 5
0 2 3 - 10
1 1 3 - 10
1 2 2 - 20

For a total of 46 possibilities. I don't see a more general formula for this.
 
The answer is the partition number of (5,3) for want of a better term which is what you've worked out, NateG, and does not follow the formula I wrote just there. Anyway, it isn't 5 choose 3. Is
\sum\binom{n}{p}\binom{n-p}{q}\binom{n-p-q}{r}

getting better? I think so.

matt
 
Last edited:
Back
Top