In how many ways can 4 balls be placed in 3 boxes if the balls are distinguishable, and the boxes are indistinguishable?

MEMEGOD Jun 16, 2024

#1**+1 **

We can use casework to solve this problem.

First, le'ts consider cases with \(4-0-0\), where all 4 balls are in one box.

Since each box is indistinguiashable, there is only 1 way to complete this.

Second, let's consider cases with \(3-1-0\)

We can calculate this by using \(4 \choose 1 \), so there are 4 ways.

Third, let's consider cases with \(2-2-0\)

There are \(4 \choose 2\) ways to complete this, so there are 6 cases.

Lastly, we have \(2-1-1\)

We have \(4 \choose 2 \) ways, to do this, so 6 additional cases.

Now, we add all of these up. We get \(1 + 4 + 6 + 6 = 17\)

So our final answer is 17.

Thanks! :)

NotThatSmart Jun 16, 2024

#2**+1 **

Sorry, that is not correct. Thanks for trying.

The correct solution is this:

For the 4, 0, 0 distribution we have only 1 way.

For the 3, 1, 0 distribution we have 4 ways.

The 2, 2, 0 distribtution is where NotThatSmart makes a mistake. It would appear to be \(4 \choose 2\) = 6 ways, but there is double counting. For example, choosing balls 1 and 3 leads to the same split as choosing balls 2 and 4. So, for the 2, 2, 0 distribution there are only 3 ways, making the total amount 14 ways.

MEMEGOD
Jun 16, 2024

#4**+1 **

I see what you mean now.

Instead of it being \(4\choose 2\), we have overcounting to consider.

Now, we only have 3 cases since there is no distinguishbility for the boxes.

Let's name the balls ball 1, 2, 3, and 4.

We have

1/2 and 3/4

1/3 and 2/4

1/4 and 2/3

We already accounted for anyother case.

Thank you for the correction, MEMEGOD!

Thanks! :)

NotThatSmart
Jun 17, 2024