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# Easy... but hard!!!

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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!!!

Apr 30, 2021

#1
+2403
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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.

Apr 30, 2021
edited by catmg  Apr 30, 2021
edited by catmg  Apr 30, 2021
#4
+124509
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Good job, catmg   !!!!

CPhill  Apr 30, 2021
#2
+124509
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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

Apr 30, 2021
#3
+2403
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What is a Stirling number?

=^._.^=

catmg  Apr 30, 2021
#5
+124509
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I'll have to admit  that I am not  familiar  with  these....but....these two websites  helped me  answer  this question

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

https://keisan.casio.com/exec/system/1292214964#!

CPhill  Apr 30, 2021
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oh no, those formulas do not look fun to use/memorize.

I shall try to study them.

=^._.^=

catmg  Apr 30, 2021
#6
+137
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OMG

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

Apr 30, 2021
#8
+2403
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Aww thank you. :))

=^._.^=

catmg  Apr 30, 2021
#9
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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. :)

Apr 30, 2021
#10
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Excellent, CentsLord   !!!!!

CPhill  Apr 30, 2021