# Counting problem.

1. Feb 27, 2004

### gimpy

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?

2. Feb 27, 2004

Staff Emeritus
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.

3. Feb 27, 2004

### Hurkyl

Staff Emeritus
Multiple objects can fit in a box, can't they?

4. Feb 27, 2004

### Loren Booda

Unless they are very, very big.

5. Feb 27, 2004

### gimpy

Yes, multiple objects can fit into a box.

I think my answer is correct, isn't it?

6. Feb 27, 2004

### matt grime

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.

7. Feb 27, 2004

### NateTG

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.

8. Feb 27, 2004

### matt grime

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: Feb 27, 2004