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(a) In how many ways can you distribute 4 distinguishable balls among 3 indistinguishable boxes?

 

(b) In how many ways can you distribute 4 indistinguishable balls among 5 distinguishable boxes?

 Jan 12, 2023
 #1
avatar+2522 
+1

a)

 

1, 1, 2

4, 0, 0

3, 1, 0

 

\(\color{brown}\boxed{3}\) cases

 Jan 12, 2023
 #4
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This is the case of 4 DISTINCT balls into 3 IDENTICAL boxes with no restrictions.

 

sumfor(i, 0, 3,(4 nCr i))==[1 + 4 + 6 + 4] ==15 ways

Guest Jan 12, 2023
 #2
avatar+2522 
+1

b) 

 

0, 0, 0, 0, 4

3, 1, 0, 0, 0

2, 1, 1, 0, 0

1, 1, 1, 1, 0

 

For the first case, there are \({5 \choose 1} = 5\) cases

 

For the second one, there are \(5 \times {4 \choose 1}= 20\) cases

 

For the third one, there are \(5 \times {4 \choose 2} = 30\) cases

 

For the final one, there are 5 cases. 

 

So, there are a total of \(5 + 20 + 30 + 5 = \color{brown}\boxed{60}\) cases.

 Jan 12, 2023
 #3
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BuilderBoy:

 

b)  You forgot one arrangement:

 

(2, 2, 0, 0, 0) ==5! / 3!2! ==10

 

So: 60 + 10 ==70 ways.

 

This is the case of 4 IDENTICAL balls into 5 DISTINCT boxes without restrictions:

 

Formula: [4 + 5 -1] nCr 4 ==8 nCr 4 ==70 ways

 

Note: To get the correct distribution, look up this page (at the very bottom of it), where there is a table made up by an expert mathematician by the name of Bogart here in Wikipedia:

 

https://en.wikipedia.org/wiki/Twelvefold_way

 Jan 12, 2023

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