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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+1952 
+1

We can use casework to solve this problem. 

 

First, le'ts consider cases with 400, 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 310

We can calculate this by using (41), so there are 4 ways. 

 

Third, let's consider cases with 220

There are (42) ways to complete this, so there are 6 cases. 

 

Lastly, we have 211

We have (42) 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 (42) = 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+1952 
+1

I see what you mean now. 

Instead of it being (42), 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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