How many ways can you partition 10 identical balls into 3 identical boxes?

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Homework Help Overview

The problem involves determining the number of ways to partition 10 identical balls into 3 identical boxes, with the condition that up to two boxes may be empty. Participants are exploring combinatorial reasoning related to this partitioning problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants describe their approaches using combinatorial methods, specifically focusing on the placement of dividers (box walls) among the balls. There is a discussion about the calculation leading to different results (55 vs. 66) and the reasoning behind these values.

Discussion Status

Some participants are clarifying their understanding of the combinatorial setup and questioning the assumptions made in their calculations. There is an acknowledgment of differing interpretations of how to account for empty boxes and the arrangement of dividers.

Contextual Notes

Participants note that the problem allows for empty boxes, which influences the counting method. There is also a mention of a teacher's feedback regarding the expected answer, prompting further exploration of the reasoning behind the calculations.

LHC
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How many ways can you place 10 identical balls into 3 identical boxes? Note: Up to two boxes may be empty.

I approached this problem as:

Let B represent ball
Let 0 represent nothing (empty)

|box wall| 0 0 B B B B B B B B B B |box wall|

So, there must be two other box walls that must be inserted, and they can be inserted in these places:

|box wall| 0 0 B B B B B B B B B B |box wall|
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

So, that would make _{11}C _{2}=55. However, my teacher says it's supposed to be 66. Could someone please explain why? Thanks.
 
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LHC said:
How many ways can you place 10 identical balls into 3 identical boxes? Note: Up to two boxes may be empty.

I approached this problem as:

Let B represent ball
Let 0 represent nothing (empty)

|box wall| 0 0 B B B B B B B B B B |box wall|

So, there must be two other box walls that must be inserted, and they can be inserted in these places:

|box wall| 0 0 B B B B B B B B B B |box wall|
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

So, that would make _{11}C _{2}=55. However, my teacher says it's supposed to be 66. Could someone please explain why? Thanks.


It's a really interesting and hard question :smile:

Here's how I approached (using yours):

| _ 1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7 _ 8 _ 9 _ 10 _ |

11C2 when you put two lines in those dashes but do not put them in same blank (so there will always be three or two boxes)
+ 11 when you put both of them together (only one box)
 
Oh, I get it! Thanks for your help! =D
 
oops.. I worded it wrong
"(so there will always be three or two boxes)"**
"(only one box or two boxes)"**

oo well, you got it ;)
 

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