How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable but the boxes are not?

my logic: there are 5 balls and each ball can choose to go in each box or not ... so that is 3^5 which is 243. then divide by 3 to account for overcounting, which makes 81. apparently, this is the wrong solution. can someone please point me in the right direction? i will be quick to reply... thanks!!!

HighSchoolDx Apr 30, 2021

#1**+7 **

In this case, case work might just be easier.

Each number shows the number of balls in that box.

0 0 5 1 way

0 1 4 5 ways

0 2 3 10 ways

1 1 3 10 ways

1 2 2 15 ways

1+5+10+10+15 = 41

Hopefully it's 41. :))

=^._.^=

Edit: The more I think, the more unsure I hate this answer. There has to be a nicer way to solve this. Casework wasn't too bad since the numbers were small, but I don't like using it too often since calculation mistakes occur. I hope someone else also answers.

catmg Apr 30, 2021

#2**+7 **

As long as we have no restrictions ( some boxes may be empty) the number of ways =

S ( 5 , 1) + S (5, 2) + S (5,3) = 1 + 15 + 25 = 41

Where "S" represents a Stirling number of the 2nd kind

CPhill Apr 30, 2021

#6**+6 **

OMG

Thank you both for answering my question with such detailed solutions... i upvoted yourposts! :)

HighSchoolDx Apr 30, 2021

#9**+7 **

nice problem. congrats on finding your solution! here is another i may present.

theres 243 arrangements to put 5 distinguishable balls in 3 **distinguishable **boxes. nice thinking with that, though :)

in these arrangements, theres one thats being counted three times - (5, 0, 0) i.e. put all the balls in one box. leaves us with 240.

for the other 240 cases, we divide by 3! instead of 3. thus, theres 240 / 6 = 40 cases. and 40 + 1 = **41. :)**

CentsLord Apr 30, 2021