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How many ways are there to put 5 balls in 2 boxes if the balls are not distinguishable and neither are the boxes?

I was thinking (5,0), (4,1), and (3,2) for a total of 12 ways. Any help is greatly appreciated.

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    $\begingroup$ You spelled it out correctly but was wondering why you would say 12 ways though? $\endgroup$
    – Satish Ramanathan
    Dec 6, 2017 at 18:11
  • $\begingroup$ /I see what you mean. How many would there be then $\endgroup$
    – ddswsd
    Dec 6, 2017 at 18:14

1 Answer 1

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It is 3: the partitions of 5 into $\leq 2 $ parts. These are pretty much just the tuples you have listed: (5), (4,1), (3,2).

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