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

 Jun 16, 2024
 #1
avatar+1075 
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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! :)

 Jun 16, 2024
 #2
avatar+52 
+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
avatar+1075 
+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
edited by NotThatSmart  Jun 17, 2024

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