(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?
+2666 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.
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
