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# Distributions

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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
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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
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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