+0  
 
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avatar+287 

1. How many ways are there to put 4 balls into 3 boxes, given that the balls can all be distinguished but the boxes are not distinguished? (Thus, for example, putting all the balls in the first box is counted as the same outcome as putting all the balls in the second box.)

 

2. How many ways are there to put 4 balls into 3 boxes, given that the balls are not distinguished but the boxes are?

 

My work so far: I'm not sure if I know what distinguished means here. I thought the solutions might've just been 3^4 which is 81, but I'm assuming that thats if everything is here is distinguished.

 

After pondering a bit, I thought of another question. What if none of them were distinguished. Would it be 0?

 Mar 23, 2020
edited by EpicWater  Mar 23, 2020
 #1
avatar+111321 
+1

1. How many ways are there to put 4 balls into 3 boxes, given that the balls can all be distinguished but the boxes are not distinguished? (Thus, for example, putting all the balls in the first box is counted as the same outcome as putting all the balls in the second box.)

 

Balls  distiguishable   Boxes not

 

Let  k be the  number of  balls  and  n  be  the  number of  boxes

 

Assuming that we  can have empty boxes......this is a little  hard to calculate,  EW

 

I'm not too sure about this....but....I believe that it  uses something  called "Stirling Numbers of the Second Kind "

 

We  have  this

 

S(4,1)  + S(4,2)  + S(4,3)

 

  1  +   6   +     7  =      14 ways  

 

 

cool cool cool

 Mar 23, 2020
 #2
avatar+111321 
+1

 

 

1. How many ways are there to put  balls into  boxes, given that the balls are not distinguished but the boxes are?

 

This  one is easier  than the first one.....again, assuming no restrictions  [ we can have emplty boxes] 

 

The  "formula"  is

 

C ( k + n - 1 , n - 1)     where k = the number of balls  and  n  =  number of boxes

 

 

cool cool cool

 Mar 23, 2020
 #3
avatar+287 
0

I'm so sorry, CPhill. I meant to type the second one not as 1 but 2.

 Mar 23, 2020
 #4
avatar+287 
0

Also, sorry about the misrenderance. It was also supposed to be 4 and 3, and not just random blanks :(

Thanks so much for your time and effort. Also, is C in this case the choose function?

 Mar 23, 2020
edited by EpicWater  Mar 23, 2020
 #5
avatar+287 
0

Thank you for the formula, though. I appreciate it and it will help a lot. Will the formula work on all types of problems like this? e.g the 3rd one I posted.

 

edit: I don't think it will.

 Mar 23, 2020
edited by EpicWater  Mar 23, 2020
 #6
avatar+483 
+1

If you want more reference info, search up "stars and bars combinatorics" on the internet. That's the formula that CPhill used in order to reach his answer for the second problem. Stars and bars only works when the "balls" or things you want to distribute are indistinguishable, and the "boxes" or groups you want to divide those things into, are distinguishable(if it was flipped around, you need to use stirling numbers like Cphill used for the first question). 

Here's a wikipedia link to it, if you really want to learn more:

 

https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

jfan17  Mar 23, 2020

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